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THE UNIVERSITY 
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PO L@ Ol A GBR A. 


BY 


G. A. WENTWORTH, 


PROFESSOR OF MATHEMATICS IN PHILLIPS EXETER ACADEMY. 


BOSTON, U.S.A.: 
PUBLISHED BY GINN & COMPANY. 
IRR De 


Entered, according to Act of Congress, in the year 1890, by 
G. A. WENTWORTH, 


in the Office of the Librarian of Congress, at Washington- 


ALL RIGHTS RESERVED. 


TypoacraPHy BY J. S. Cusuine & Co., Boston, US8.A. 


PRESSWoORK BY GINN & Co., Boston, U.S.A. 


elec lire tan (nh: 


es 


Tus book, as the name implies, is written for High Schools and 

Academies, and is a thorough and practical treatment of the prin- 
ciples of Elementary Algebra. It covers sufficient ground for 
admission to any American college, and with the author’s Col- 
lege Algebra makes as extended a course as the time allotted 
to this study in our best schools and colleges will allow. Great 
care has been taken to present the best methods, so that students 
in going from the lower book to the higher will have a good 
foundation, and have nothing to unlearn. 

The problems are carefully graded. They are for the most part 
new ; either original or selected from recent examination papers. 
They are sufficiently varied and interesting, and are not so difficult 
as to discourage the beginner. The early chapters are quite full; 
for even if a student is perfectly familiar with the operations of 
“Arithmetic, he must have time to learn the language and the 
fundamental processes of Algebra. 

The introductory chapter should be read and discussed in the 
recitation room. This chapter’ brings before the student in brief 
review the knowledge he has already gained from the study of 
Arithmetic, states and proves the general laws of numbers, sets 
forth clearly the advantage of using letters to represent numbers 
in the statement of general laws, and leads him to see at the 
outset that Algebra, like Arithmetic, treats of numbers. In this 
chapter, also, the meaning of negative quantities is explained, and 
the laws which regulate the combinations of different arithmetical 
numbers are shown to apply to algebraic numbers. It is hoped 


1V PREFACE. 


that a free discussion of these elementary principles will do much 
to prevent that vagueness which the beginner invariably experi- 
ences if he fails to connect the laws of Algebra with what he has 
learned in Arithmetic. 

Answers to the problems are bound separately, in paper covers, 
and will be furnished free to pupils when teachers apply to the 
publishers for them. 

Any corrections or suggestions oe to the work will be 
thankfully received. ; 

G. A. WENTWORTH. 
PHILLIPS EXETER ACADEMY, 
June, 1890. 


CHAPTER 


I. 
iL}. 


COUN Ds Bane) 


INTRODUCTION . 

ADDITION AND SUBTRACTION . 
MULTIPLICATION 

DIVISION . 

SIMPLE EQUATIONS 
MULTIPLICATION AND DIvisIon . 
Factors . 

Common Factors AND MULTIPLEs . 
FRACTIONS 

FRACTIONAL EQUATIONS 


SIMULTANEOUS EQUATIONS OF THE First DEGREE 
PROBLEMS INVOLVING Two Unknown NuMBERS . 


INEQUALITIES 

InvoLuTion AND Evouurion . 
THEORY OF EXPONENTS 

RapicaAL EXPRESSIONS . 

IMAGINARY EXPRESSIONS . 
QUADRATIC EQUATIONS . 
SIMULTANEOUS QUADRATIC EQUATIONS 
PROPERTIES OF QUADRATICS . 


Ratio, PRopoRTION, AND VARIATION . 


PROGRESSIONS 

PROPERTIES OF SERIES . 
BrnoMIAL THEOREM . 
LoGARITHMS 

GENERAL Review EXERCISE. 





SCHOOL ALGEBRA, 





CHAPTER I. 
INTRODUCTION. 


1, Units. In counting separate objects the standards by 
which we count are called units; and in measuring contin- 
uous magnitudes the standards by which we measure are 
called units. Thus, in counting the boys in a school, the 
unit is a boy; in selling eggs by the dozen, the unit is a 
dozen eggs; in selling cloth by the yard, the unit is a yard 
-of cloth; in measuring short distances, the unit is an inch, 
a foot, or a yard; in measuring long distances, the unit is 
a rod or a mile. 


2. Numbers. Repetitions of the unit are expressed by 
numbers. Ifa man, in sawing logs into boards, wishes to 
keep a count of the logs, he makes a straight mark for 
every log sawed, and his record at different times will be 
as follows: 


ee ee il ae dN aL 
me ee IN Le. LINN 


These representative groups are named one, two, three, 
four, five, six, seven, eight, nine, ten, etc., and are known 
collectively under the general name of numbers, It is 
obvious that these representative groups will have the 
same meaning, whatever the nature of the unit counted. 


2 SCHOOL ALGEBRA. 


3. Quantities. The word “quantity” (from the Latin 
guantus, how much) implies both a unit and a number. 
Thus, if we inquire how much wheat a bin will hold, we 
mean how many bushels of wheat it will hold. If we 
inquire how much carpeting there is in a certain roll, 
we mean how many yards of carpeting. If we inquire 
how much wood there is on a certain wood-lot, we mean 
how many cords of wood. 


4, Number-Symbols in Arithmetic. Instead of groups of 
straight marks, we use in Arithmetic the arbitrary sym- 
bols 1, 2, 3, 4, 5, 6, 7, 8, 9, called figures, for the numbers 
one, two, three, four, five, six, seven, erght, nine. 

The next number, ten, is indicated by writing the figure 
1 in a different position, so that it shall signify not one, but 
ten. This change of position is effected by introducing a 
new symbol, 0, called nought or zero, and signifying none. 
Thus, in the symbol 10, the figure 1 occupying the second 
place from the right, signifies a collection of ten things, and 
the zero signifies that there are no single things over. The 
symbol 11 denotes a collection of ten things and one thing 
besides. All succeeding numbers up to the number con- 
sisting of 10 tens are expressed by writing the figure for 
the number of tens they contain in the second place from 
the right, and the figure for the number of units besides in 
the first place. The number consisting of 10 tens is called 
a hundred, and the hundreds of a number are written in the 
third place from the right. The number consisting of 10 
hundreds is called a thousand, and the thousands are writ- 
ten in the fourth place from the right; and so on. 


5. The Natural Series of Numbers. Beginning with the 
number one, each succeeding number is obtained by put- 
ting one more with the preceding number. If from a given 


INTRODUCTION. 3) 


point marked 0, we draw a straight line to the right, and 
beginning from this point lay off units of length, the suc- 
cessive repetitions of the unit will be denoted by the natural 
series of numbers 1, 2, 3, 4, etc. Thus, 


i: 2 3 4 5 6 u ete. 





0 


6. The reader will notice that number symbols in Arith- 
metic stand for particular numbers, and that these symbols 
indicate a method of making up the number, but not neces- 
sarily the method by which the number is actually made 
up. Thus, if a man has 66 dollars in bank-notes, he may 
have, as the number 66 indicates, 6 ten-dollar bills and 6 
one-dollar bills, but this is not the only way in which the 
66 dollars may be made up. 


7. Integral and Fractional Numbers. When the things 
counted are whole units, the numbers which count them 
are called whole numbers, integral numbers, or integers, where 
the adjective is transferred from the things counted to the 
numbers which count them. But if the things counted are 
only parts of units, the numbers which count them are 
called fractional numbers, or simply fractions, where again 
the adjective is transferred from the things counted to the 
numbers which count them. 

To represent the parts of a given unit, two number- 
symbols are used, one to name the parts into which the 
unit is divided, and therefore called the denominator, and 
the other to denote the number of parts taken, and there- 
fore called the numerator. The denominator is written 
below the numerator with a line between them. ‘Thus, in 
the fraction ~ the 9 shows that the unit is divided into 
nine equal parts, called ninths of the unit, and the 7 shows 
that seven of these equal parts are taken. 


4 SCHOOL ALGEBRA. 


8. Principal Signs of Operations. The sign +, read plus, 
indicates that the number after the sign is to be added to 
the number before the sign. Thus, 5+ 4 means that 4 
is to be added to 5. 

The sign —, read manus, indicates that the number after 
the sign is to be subtracted from the number before the sign. 
Thus, 8—4 means that 4 is to be subtracted from 8. 

The sign X, read tzmes or ina, indicates that the number 
after the sign is to be multiplied by the number before the 
sign. Thus, 5x4 means that 4 is to be multiplied by 5. 

The sign +, read divided by, indicates that the number 
before the sign is to be dided by the number after the 
sign. Thus, 8+4 means that 8 is to be divided by 4. 

The operation of division is also indicated by placing 
the dividend over the divisor with a line between them. 
Thus, $ means the same as 8+ 4. 


9, Signs of Relation. The sign =, read equals, or 1s equal 
to, when placed between two numbers, indicates that they are 
equal. Thus, 8-+4—12 means that 8+4 is the same as 12. 

7 The sign >, read 2s greater than, indicates that the num- 
ber which precedes the sign is greater than the number 
which follows it. Thus, g +4> 10 means that 8+ 41s 
greater than 10. 

The sign <, read 2s less than, indicates that the number 
which precedes the sign is less than the number which fol- 
lows it. Thus, 8+4<16 means that 8+4 is less than 16. 


10. Signs of Deduction and of Continuation. The sign .° 
stands for the word “therefore” or “hence.’’ The sign 
tease or —-———~— stands for the words ‘“‘and so on.” 


11, Number-Symbols in Algebra. Algebra, like Arith- 
metic, treats of numbers, and employs the letters of the 
alphabet in addition to the figures of Arithmetic to represent 


| 


INTRODUCTION. 5 


numbers. The letters of the alphabet are used as general 
symbols of numbers to which any particular values may be 
assigned. In any particular problem, however, a letter must 
be supposed to have the same particular value throughout 
the investigation or discussion of the problem. 

These general symbols are of great advantage in investi- 
gating and stating general laws; in exhibiting the actual 
method in which a number is mene up; and in represent- 
ing unknown numbers which are to be discovered from their 
relations to known numbers. 

The advantage of representing numbers by letters will 
be more clearly seen later on. For the present it will be 
sufficient for the beginner to understand that every letter, 
and every combination of letters, and every combination of 
Jigures and letters used in Algebra, represents some number. 
Thus, the number of dollars in a package of bank-notes 
can be represented by x; but if the package consists of ten- 
dollar bills, five-dollar pale two-dollar bills, and one-dollar 
bills, and if we denote the number of ten-dollar bills by a, 
of five-dollar bills by 4, of two-dollar bills by e, and of one- 
dollar bills by d, the whole number of dollars in the package 
will be represented by 10a+56+2c+d. | 

In this particular case x and 10a+56+42c-+d both 


stand for the same number. 


12, Substitution. It is obvious that the same operation 
on each of the above expressions will produce results that 
agree in value, and therefore that either may be substituted 
for the other at pleasure. In short, 

EKvery algebraic expression represents some number, and 
may be operated upon as tf it were a single symbol standing 
for the number which rt represents. 


13, Factors. When a number consists of the product of two ° 
or more numbers, each of these numbers is called a factor of 


6 SCHOOL ALGEBRA. 


the product. If these numbers are denoted by letters, the 
sign X 1s omitted. Thus, instead of a x b, we write ab. 


14, Coefficients) A known factor prefixed to another 
factor to show how many times that factor is taken is called 
a coefficient, 


15. Powers. A product consisting of two or more equal 
factors is called a power of that factor. | 

The index or exponent of a power is a small figure placed 
at the right of a number, to show how many times the 
number is taken as a factor. Thus, 2‘ is written instead of 
2xX2x2~x 2; a’ instead of aaa. 

The second power of a number is generally called the 
square of that number; the third power of a number, the 
cube of that number. 


16. Roots. The root of a number is one of the equal fac- 
tors of that-number; the square root of a number is one 
of the two equal factors of that number; the cube root of a 
number is one of the three equal factors of that number ; 
and so on. The sign V, called the radical sign, indicates 
that a root is to be found. Thus, V4, or V4, means that 
the square root of 4 is to be taken; 8 means that the 
cube root of 8 is to be taken; and so on. 

The figure written above the radical sign is called the 
index of the root. 


17. An algebraic expression is a number written with alge- 
braic symbols; an algebraic expression consists of one sym- 
bol, or of several symbols connected by signs of operation. 

A term is an algebraic expression the parts of which are 
not separated by the sign of addition or subtraction. Thus, 
3ab, 52x 4y,3ab+4zy are terms. 3 


INTRODUCTION. 4 


A simple expression is an expression of one term. 
A compound expression is an expression of two or more 
terms. 


18. Positive and Negative Terms. The terms of a com- 
pound expression preceded by the sign + are called posi- 
tive terms, and the terms preceded by the ‘sign — are called 
negative terms. The sign + before the first term is omitted. 


19, Parentheses. If a compound expression is to be 
treated as a whole it is enclosed in a parenthesis. Thus, 
2x (10-++ 5) means that we are to add 5 to 10 and multiply 
- the result by 2; if we were to omit the parenthesis and 
write 2x 10+ 5, the meaning would be that we were to 
multiply 10 by 2 and add 5 to the result. 

Instead of parentheses, we use with the same meaning 
brackets [ ], braces §?, and a straight line called a vinculum. 


5 
Thus, (5+ 2), [5+ 2], {5+ 2} Wet, i of all mean that 
the expression 5+ 2 is to be treated as the single symbol 7. 


20. Rules for removing Parentheses. If a man has 10 dol- 
lars and afterwards collects 8 dollars and then 2 dollars, 
it makes no difference whether he adds the 8 dollars to his 
10 dollars, and then the 2 dollars, or puts the 3 and 2 
dollars together and adds their sum to his 10 dollars. 

The first process is represented by 10-++3-+2. 

The second process is represented by 10-+(8 + 2). 


Hence 10+(3+2)=104+38+42. (1) 


If a man has 10 dollars and afterwards collects 3 dol- 
lars and then pays a bill of 2 dollars, it makes no differ- 
ence whether he adds the 38 dollars collected to his 10 
dollars and pays out of this sum his bill of 2 dollars, or 
pays the 2 dollars from the 38 dollars collected and adds 
the remainder to his 10 dollars. 


8 SCHOOL ALGEBRA. 


The first process is represented by 10+ 38— 2. 
The second process is represented by 10 + (3 — 2). 


Hence 10 + (8 — 2)=10+3—2. (2) 


From (1) and (2) it follows that if a compound expres- 
sion is to be added, the parenthesis may be removed and 
each term in the parenthesis retain its prefixed sign. 

If a man has 10 dollars and has to pay two bills, one of 
3 dollars and one of 2 dollars, it makes no difference whether 
he takes 3 dollars and 2 dollars in succession, or takes the 3 
and 2 dollars at one time, from his 10 dollars. 

The first process is represented by 10 —3— 2. 

The second process is represented by 10—(38+ 2). 


Hence 10—(8+2)=10—3— 2. (3) 


If a man has 10 dollars consisting of 2 five-dollar bills, 
and has a debt of 3 dollars to pay, he can pay his debt by 
giving a five-dollar bill and receiving 2 dollars. 

This process is represented by 10 —5-+ 2. 

Since the debt paid is three dollars, that is, (6—2) dol- 
lars, the number of dollars he has left can evidently be 


expressed by 10 — (5 ~ 2) 
Hence 10 — (5 — 2) = 10—5+-2. (4) 


From (8) and (4) it follows that if a compound expres- 
sion is to be subtracted, the parenthesis may be removed, 
provided the sign before each term within the parenthesis 
is changed, the sign + to —, and the sign — to +. 


Exercise 1. 
Perform the operations indicated, and simplify . 
1. 7+(38—2) 4. 5x(2+8). 7. (7—8)x (6—2). 
2. 7—(8--2). 5. (564+8)+2. 8 (8—2)+(5—2). 
38. 7—(842). 6. 5x(8—2) 9. 3x(12—6—2). 


INTRODUCTION. 9 


21, Fundamental Laws of Numbers. We are so occupied 
in Arithmetic with the application of numbers to the ordi- 
nary problems of every-day life that we pay little attention 
to the investigation of the fundamental laws of numbers. 
It is, however, very important that the beginner in Algebra 
should have clear ideas of these laws, and of the extended 
meaning which it is necessary to give in Algebra to cer- 
tain words and signs used in Arithmetic; and that he 
should see that every such extension of meaning is con- 
sistent with the meaning previously attached to the word 
or sign, and with the general laws of numbers. We shall, 
therefore, give general definitions for the fundamental oper- 
ations upon numbers and then state the laws which apply 
to them. 


. 22, Addition. The process of finding the result when 
two or more numbers are taken together is called addition, 
and the result is called the sum, 


23. Subtraction, The process of finding the result when 
one number is taken from another is called subtraction, and 
the result is called the difference or remainder. The number 
taken away is called the subtrahend, and the number from 
which the subtrahend is taken is called the minuend. 

In practice the difference is found by discovering the 
number which must be added to the subtrahend to give the 
minuend. Ifthe subtrahend consists of two or more terms, 
we add these terms and then determine the number which 
must be added to their sum to make it equal to the minu- 
end. Thus, if a clerk in a store sells articles for 10 cents, 
15 cents, and 380 cents, and receives a dollar bill in pay- 
ment, he makes change by adding these items and then 
adding to their sum enough change to make a dollar. 

From the nature of this process it 1s obvious that the 
general laws of numbers which apply to addition apply 


10 SCHOOL ALGEBRA. 


also to subtraction, and that we may take for the general 
definition of subtraction 


The operation of finding from two given numbers, called 
minuend and subtrahend, a third number, called difference, 
which added to the subtrahend will gwe the minuend. 


24, Multiplication. The process of finding the result 
when a given number is taken as many times as there are 
units in another number is called multiplication, and the re- 
sult is called the product, 

This definition fails when the multiplier is a fraction, for 
we cannot take the multiplicand a fraction of a tume. We 
therefore consider what extension of the meaning of multi- 
plication can be made so as to cover the case in qt: ““:on. 
When we multiply by a fraction we divide the multiplicand 
into as many equal parts as there are units in the denomi- 
nator and take as many of these parts as there are units 
in the numerator. If, for instance, we multiply 8 by ?, we 
divide 8 into four equal parts and take three of these parts, 
getting 6 for the product. We see that 7? is ¢ of 1, and 6 
is } of 8; that is, the product 6 is obtained from the mul- 
tiplicand 8 precisely as the multiplier $ is obtained from 1. 
The same is true when the multiplier is an integral number. 


Thus, in 6 X 8 = 48, 
the multiplier 6is 1+14+14+141-41, 
and the product 48 is 8+8+8+8+48-+8. 


We may, therefore, take for. the general definition of 
multiplication 


The operation of finding from two given numbers, called 
multiplicand and multipler, a third number called product, 
. which is formed from the multiplicand as the multiplier ts 
formed from unity. 


INTRODUCTION. ll 


25. Division. To divide 48 by 8 is to find the number 
of times it 1s necessary to take 8 to make 48. Here the 
product and one factor are given and the other factor is 
required. We may therefore take for the general definition 
of division 

The operation by which when the product and one factor 
are given the other factor is found. 


With reference to this operation the product is called 
the dividend, the given factor the divisor, and the required 
factor the quotient. 


26. The Commutative Law. If we have a group of 3 
- things and another group of 4 things, we shall have a 
group of 7 things, whether we put the 3 things with the 
4 things or the 4 things with the 8 things; that is, 


4+38=3-4-4. 


It is evident that the truth of the above statement does 
not depend upon the particular numbers 8 and 4, but that 
the statement is true for any two numbers whatever. Thus, 
in case of any two numbers we shall have 


First number + second number = second number + first 
number. 


If we let a stand for the first number and 6 for the second 
number, this statement may be written in the much shorter 
form 

atb=b-+a. 

This is the commutative law of addition, and may be 
stated as follows: 

Additions may be performed in any order. 


27. Also, if we have 5 lines of dots with 10 dots in a 
line, the whole number of dots will be expressed by 5 x 10. 


oy SCHOOL ALGEBRA. 


If we consider the dots as 10 columns with 5 dots in a 
column, the number will be expressed by 10 x 5. 


That is, Dx 1G == 102. 
Again, if we divide a given length into 6 equal parts, 
3 2 





one-third of the line will contain 2 of these parts, and one- 
half the line will contain 3 of these parts. Now one-third 
of one-half will be 1 of these parts, and one-half of one- 
third will be 1 of these parts; that is, 


txt=3Xt. 
Therefore, if a and 6 stand for any two numbers, integral 
or fractional, we shall have 
ae 
This is the commutative law of multiplication, and may 
be stated as follows: 


Multipheations may be performed in any order. 


28. The Distributive Law. The expression 4 x (5+ 38) 
means that we are to take the sum of the numbers 5 and 3 
four times. The process can be represented by placing five 
dots in a line, and a little to the right three more dots in 
the same line, and then placing a second, third, and fourth 
line of dots underneath the first line and exactly similar to 


it. eoee0 eeo0e 
eooeo0oe5e eee 
ecec0e0ne eee 
eooe50e eee 


INTRODUCTION. 183 


There are (5+ 3) dots in each line, and 4 lines. The 
total number of dots, therefore, is 4 x (5+ 8). 

We see that in the left-hand group there are 4 x 5 dots, 
and in the right-hand group 4 x 3 dots. The sum of these 
two numbers (4 x 5)+(4 x 3) must be equal to the total 
number; that is, 


4x (8+38)=(4x 5)+ (4 8). 


Again, the expression 4 x (8 — 3) means that 3 is to be 
taken from 8, and the remainder to be multiplied by 4. 
The process can be represented by placing eight dots in a 
line and crossing the last three, and then placing a second, 
third, and fourth line of dots underneath the first line and 
exactly similar to it. 


@®@e@38 @ 
eee @ 
e@e8 @ 
e@e38 @ 
@®@e @ 
Se OQ “Oe “B. 
e ee ea @ 
eB C8 eA OQ 


The whole number of dots not crossed in each line is 
evidently (8 —3), and the whole number of lines is 4. 
Therefore the total number of dots not crossed is 


Ao 3). 

The total number of dots (crossed and not crossed) is 
(4 x 8), and the total number of dots crossed is (4 X 3). 
Therefore the total number of dots not crossed is 

(4 x 8)—(4x 8); 
that is, 4x (8—3)=(4 x 8)—(4 x 8). 
Hence, by the commutative law 
(8 —38)x4=(8X 4)—(8 x 4). 
~ » In like manner, if a, 6, ¢, and d stand for any numbers, 


“wehave = gx (b4¢—d)=ab+ ac—ad, 


: 14 SCHOOL ALGEBRA. 


This is the distributive law, and may be stated as follows: 


In multiplying a compound expression by a simple ex- 
pression the result 1s obtained by multyplying each term of 
the compound expression by the simple expression, and writ- 
mg down the successwe products with the same signs as 
those of the original terms. 


29. The Associative Law. The terms of an expression may 
be grouped in any manner. For if we have several num- 
bers to be added, the result will evidently be the same, 
whether we add the numbers in succession or arrange them 
in groups and add the sums of these groups. Thus, 


atbtetdte 
= tet A OeT 2) Ad ce) 
=(a+6)+(e+d+e). 

Likewise, if in the rectangular solid represented in the 
margin we suppose AB to contain 5 units of length, BC 3 
units, and CD 7 units. The base may 
be divided into square units. There 
will be 3 rows of 5 square units each. 
Upon each square unit a cubic unit may 
be formed, and we shall have (3x5) 
cubic units. Upon these another tier of 
(3 x 5) cubic units may be formed, and 
then another tier of the same number, 
and the process continued. until we have 7 tiers of (3 x 5) 
cubic units. Hence the number of cubic units in the solid 
will be represented by 7 x (8 X 5). 

Upon the right-hand square in the back row a pile of 7 
cubic units may be formed, upon the next square to the 
left another pile of 7 cubic units may be formed, and 
upon the next square another, and the process continued 
until we have a pile of 7 cubic units on each square in the 


Ly 


D 
4 
, 
a 
é 
4 
4 





INTRODUCTION. 15 


back row. We shall then have (5 x 7) cubic units in the back 
tier, and as we can have 3 such tiers, the number of cubic 
units in the solid will be represented by 3 x (5 x 7). 

Again, if we form a pile of 7 cubic units on the right-hand 
square of the back row, then another pile of 7 cubic units on 
the next square in front, another pile of 7 cubic units on the 
next square in front, we shall have a tier of (3 x 7) cubic 
units. We can have 5 such tiers, and the number of cubic 
units in the solid will now be represented by 5 x (8 x 7). 

It follows, therefore, that the total number of cubic units 
in the solid may be represented by 

7x(8x 5), or by 3X(5X 7), or by 5X (8X 7). 

It is obvious that no part of this proof depends upon the 
particular numbers 38, 5, and 7, but the law holds for any 
arithmetical numbers whatever, and may be expressed by 

COG a0) = a DEC OXI re) 

This is called the associative law of addition and multi- 
plication, and may be stated as follows: . 

The terms of an expression, or the factors of a product, 
may be grouped in any manner. 


30. The Index Law. 
Since a= a0, and a = aaa, 
OC 0G On a a 
Ce ate CO a= = 
If a stands for any number, and m and » for any integers, 
since a™”= aaa--- to m factors, 
and a" = aaa to ” factors, 
a” X a” = (aaa ----- to m factors) X (aaa. ..... to n factors), 
= aaa: to (m+n) factors, 


es poe 


— 


16 SCHOOL ALGEBRA. 


Hence, the index law may be stated as follows: 


The index of the product of two powers of the same number 
as equal to the sum of the ndices of the factors. 


31. These four laws, the commutative, the distributive, 
the associative, and the index laws, are the fundamental 
laws of Arithmetic, and together with the daw of signs, 
which ‘will be explained hereafter, they constitute the 
fundamental laws of Algebra. 


32, Quantities Opposite in Kind. If a man gains 6 dollars 
and then-loses 4 dollars, his actual gain, or, as we com- 
monly say, his ne? gain, is 2 dollars; that is, 4 dollars’ loss 
cancels 4 dollars of the 6 dollars’ gain and leaves 2 dollars’ 
gain. If he gained 6 dollars and then lost 6 dollars, the 6 
dollars’ loss cancels the 6 dollars’ gain, and his net gavn is 
nothing. If he gained 6 dollars and then lost 9 dollars, 
the 6 dollars’ gain cancels 6 dollars of the 9 dollars’ loss, 
and his net loss is 3 dollars. In other words, loss and gain 
are quantities so related that one cancels the other wholly 
or in part. 

If the mercury in a thermometer rises 12 degrees and 
then falls 7 degrees, the fall of 7 degrees cancels 7 degrees 
of the 7vse, and the net rise is 5 degrees. If it rises 12 de- 
grees and then falls 12 degrees, the net rise is nothing. 
If it ruses 12 degrees and fads 15 degrees, there is a net 
fall of 3 degrees. In other words, rise and fal are quan- 
tities so related that one. cancels the other wholly or in 
part. 

An opposition of this kind also exists in motion forwards 
and motion Jdackwards; in distances measured east and 
distances measured west; in distances measured north and 
distances. measured Sones in assets and debts; in time be- 
Jore and time after a fied date; and so on. 


INTRODUCTION. let fe 


83, Algebraic Numbers. Jf we wish to add 4 to 3, we 


begin at 4 in the natural series of numbers, 
OAR er Poe Oe Gh) oa oO 


count 3 units forwards, and arrive at 7,the sum sought. If 
we wish to subtract 3 from 7, we begin at 7 in the natural 
series of numbers, count 3 units backwards, and arrive at 4, 
the difference sought. If we wish to subtract 7 from 7, we 
begin at 7, count 7 units backwards, and arrive at 0. If 
we wish to subtract 7 from 4, we cannot do it, because 
when we have counted backwards as far as 0 the natural 
serves of numbers comes to an end. 

In order to subtract a greater number from a smaller it 
is necessary to asswme a new series of numbers, beginning 
at zero and extending to the left of zero. The series to the 
left of zero must proceed from zero by the repetitions of the 
unit, precisely like the natural series to the right of zero; 
and the opposition between the right-hand series and the 
left-hand series must be clearly marked. This opposition 
is indicated by calling every number in the right-hand 
series a positive number, and prefixing to it, when written, 
the sign +; and by calling every number in the left-hand 
series a negative number, and prefixing to it the sign —. 
The two series of numbers will be written thus: 


soeee eee Bea UM ay Wart L Ge a 2 Sea ra asa 
— a Se ee ee) eee ee ee ee 


ALGEBRAIC SERIES OF NUMBERS. 


If, now, we wish to subtract 9 from 6, we begin at 6 in 
the positive series, count 9 units in the negative direction 
(to the left), and arrive at — 3 in the negative series; that 
is, 6B—9=—83. 

The result obtained by subtracting a greater number from 
a less, when both are positive, is always a negative number. 


18 SCHOOL ALGEBRA. 


In general, if a and 6 represent any two numbers of the 
positive series, the expression a—6 will be a positive num- 
ber when a is greater than 0; will be zero when a is equal 
to 6; will be a negative number when a is less than 0. 

In counting from left to right in the algebraic series num- 
bers wmcrease in magnitude; in counting from right to left 
numbers decrease in magnitude. Thus —3, —1, 0, +2, 
+4 are arranged in ascending order of magnitude. 


34, We may illustrate the use of algebraic numbers as 

follows: Wat An dts 8 20 
-+-——— 
va aa eel C 

Suppose a person starting at A walks 20 feet to the right 
of A, and then returns 12 feet, where will he be? Answer: 
At C, a point 8 feet to the right of A; that is, 20 feet —12 
feet = 8 feet; or, 20 —12=8. 

Again, suppose he walks from A to the right 20 feet, and 
then returns 20 feet, where will he be? Answer: At A, 
the point from which he started; that is, 20 — 20 = 0. 

Again, suppose he walks from A to the right 20 feet, and 
then returns 25 feet, where will he now be? Answer: At 
D, a point 5 feet to the left of A; that is, 20 —25—=— 5; 
and the phrase ‘5 feet to the left of A’’ is now expressed 
by the negative quantity, — 5 feet. 


35, Every algebraic number, as + 4 or —4, consists of a 
sign + or — and the absolute value of the number. The 
sign shows whether the number belongs to the positive or 
negative series of numbers; the absolute value shows what 
place the number has in the positive or negative series. 

When no sign stands before a number, the sign + is 
always understood. Thus 4 means the same as +4, a 
means the same as-+a. But the sign — is never omitted. 


INTRODUCTION. 19 


86. Two algebraic numbers which have, one the sign + 
and the other the sign —, are said to have unlike signs. 

Two algebraic numbers which have the same absolute 
values, but unlike signs, always cancel each other when 


combined. Thus +4—4=0, +a—a=0. 


87. Double Meanings of the Signs + and —. The use of 
the signs + and — to indicate addition and subtraction 
must be carefully distinguished from the use of the signs + 
and — to indicate in which series, the positive or the nega- 
tive, a given number belongs. In the first sense they are 
signs of operations, and are common to Arithmetic and 
Algebra; in the second sense they are signs of opposition, 
and are employed in Algebra alone. 


38, Addition and Subtraction of Algebraic Numbers. An 
algebraic number which is to be added or subtracted is 
often inclosed in a parenthesis, in order that the signs + 
and —, which are used to distinguish positive and negative 
numbers, may not be confounded with the + and — signs 
that. denote the operations of addition and subtraction. 
Thus + 4-++ (— 8) expresses the sum, and + 4 — (— 8) ex- 
presses the difference, of the numbers ++ 4 and — 3. 

In order to add two algebraic numbers we begin at the 
place in the series which the first number occupies and 
count, wn the direction indicated by the sign of the second 
number, as many units as there are in the absolute value 
of the second number. ) 

Thus the sum of +4-+ (4+ 38) is found by counting from 
+4 three units in the positive direction; that is, to the 
right, and is, therefore, ++ 7. 

The sum of + 4-+(— 8) is found by counting from + 4 
three units in the negative direction; that is, to the left, and 
is, therefore, + 1. 


20 SCHOOL ALGEBRA. 


The sum of —4-+(+8) is found by counting from — 4 
three units in the positive direction, and\is, therefore, — 1. 


Ne Dae SnD ae Lal Oy elem 9 eee TA se 
a Se Se Ce Oe ee ee ee eeEeEE———e—EEE—EE EE 


The sum of —4-+(— 8) is found by counting from — 4 
three units in the negative direction, and is, therefore, — 7. 

Hence to add two or more algebraic numbers, we have 
the following rules: 


Case I. When the numbers have like signs. Find the 
sum of their absolute values, and prefix the common sign to 
the result. 


CasE II. When there are two numbers with wnlike signs. 
Find the difference of their absolute values, and ag to the 
result the sign of the greater number. 


Case III. When there are more than two numbers with 
unlike signs. Combine the first two numbers and this result 
with the third number, and so on; or, find the sum of the 
positwe numbers and the sum of the negative numbers, take 
the difference between the absolute values of these two sums, 
and prefix to the result the sign of the greater sum. 


39. The result is called the sum. It is often called the 
algebraic sum, to distinguish it from the arithmetical sum, 
that is, the sum of the absolute values of the numbers. 


40, Subtraction. In order to subtract one algebraic num- 
ber from another, we begin at the place in the series which 
the minuend occupies and count in the direction opposite to 
that indicated by the sign of the subtrahend as many units 
as there are in the absolute value of the subtrahend. 

Thus, the result of subtracting + 3 from + 4 is found by 
counting from +4 three units in the negatwe direction; 
that is, in the direction opposite to that indicated by the sign 
+ before 3, and is, therefore, +1. 


INTRODUCTION. _ 21 


The result of subtracting —3 from +4 is found by count- 
ing from +4 three units in the positive direction; that is, 
in the direction opposite to that indicated by the sign — be- 
fore 3, and is, therefore, ++ 7. 

The result of subtracting +3 from —4 is found by count- 
ing from —4 three units in the negative SE a and is, 
therefore, — 7. 

The result of subtracting —3 from —4 is found by count- 
ing from —4 three units in the positive direction, and is, 
therefore, — 1. 

Collecting the results obtained in addition and subtrac- 
tion, we have | 


ADDITION. SUBTRACTION. 
444(—3)=44—-8=41. 44—(43)=4+4-38=41. 
+44(43)=4+443=+7. +4-(-3)=4+44+3=4+7. 
—44(—8)=—4-8=—7.  —4—(48)=—4—3=~—7,. 
—4+(+3)=—4+38=—-1. —4—(—3)=—4438=— 


No part of this proof depends upon the particular num- 
bers 4 and 8, and hence we may employ the general symbols 
a and 6 to represent the absolute values of any two alge- 
braic numbers. We shall then have 


ADDITION. SUBTRACTION. 
+a+(—b)=+a—b. +a—(+6)=+a—b6. (1) 
+at+(+6)=+a+0b. -a—(-d)=Ta-Fb (2) 
—a+(—b)=—a—b. —a—(+6)=—a—b. (3) 
—a+(+6)=—a+b. —a—(—b)=—a+b. (4) 


From (1) and (8), it is seen that "subtracting a positive 
wumber is equivalent to adding an equal negative number. 

From (2) and (4), it is seen that subtracting a negative 
number is equivalent to adding an equal positive number. 


22, SCHOOL ALGEBRA. 


To subtract one algebraic number from another, we have, 
‘ therefore, the following rule: 


Change the sign of the subtrahend, and rides the subtra- 
hend to the minuend. 


This rule is consistent with the definition of subtraction 
given in § 23;. for, if we have to subtract — 4 from +3, we 
must add +4 to the subtrahend —4 to cancel it, and then 
add + 3 to obtain the minuend; that is, we must add +7 
to the subtrahend to get the minuend, but + 7 is obtained 
by changing the sign of the subtrahend —4, making it +4, 
and adding it to +38, the minuend. 


41, The commutative law of addition applies to algebraic 
numbers, for +4-+(— 3)=—8+(+4). In the first case 
we begin at +4 in the series, count three units to the left, 
and arrive at +-1; in the second case we begin at —3 in 
the series, count four units to the right, and arrive at +1. 

The associative law, also, of addition is easily seen to 
apply to algebraic numbers. 


42. Multiplication and Division of Algebraic Numbers. By 
the definition of multiplication, § 24. 


Since +3=+1+1+1; 
“3X (4+8)=+8+8+8 
= -+ 24, 
and 3 x (—8)=—8—8—8 
= — 24 
Again, since —8=—1—1-1; 
..(—3)xX 8=—8—8-—8 
= — 24, 
and ohio, (+58) x¢ (8) eee) (8) aa) 
=+8148+8 


= + 24, 


INTRODUCTION. Os 


No part of this proof depends upon the particular num- 
bers 8 and 8. If we use a to represent the absolute value 
of any number, and 6 to represent the absolute value of 
any other number, we shall have 


(+a)x (4+ 6)=+ a8. (1) 
(+ a) x (— 6) =— ab. (2) 
(+a) x GC b)=— ab. (3) 
(—a)xX(— b)=+ ab. (4) 


43, Law of Signs in Multiplication, From these four cases 
it follows that, in finding the product of two algebraic 
numbers, 


TInke signs gwe +,and unlike signs give —. 


44, Law of Signs in Division. 


Since (+a) xX (+0)=+ ad, .«. +ab+(+a)=+0. 
Since (+a) x(—b)=—ab, «. —ab+(+a)=—20. 
Since (—a)X(+6)=—ab, .. ~ab+(—a)=+0. 
Since (—a) X (—b)=+ ad, «. +ab+(—a)=—8. 


That is, if the dividend and divisor have like signs, the 
quotient has the sign-++; and if they have unlike signs, 
the quotient has the sign —. Hence, in division, 


TInke signs gwe +; unlike signs giwe —. 


45. From the four cases of multiplication that we have 
given in § 42 it will be seen that the absolute value of 
each product is independent of the signs, and that the signs 
are independent of the order of the factors. Hence the com- 
mutative and associative laws of multiplication hold for all 
algebraic numbers. 


94 SCHOOL ALGEBRA. 


46. The distributive law also holds; for, if 
a(b+c)=ab-+ae, 
then —a(b+c)=— ab—ace, 
and (6+ c)(— a) = 6b (—a) + e(—a). 
Therefore, for all values of a, 6, and e, 
a(b+¢)=ab+ae. 


From the nature of division the distributive law which 
applies to multiplication applies also to division. 


47, We have now considered the fundamental laws of 
Algebra, and for convenience of reference we formulate 
them below: | 

a+(bt+c)=a+b+e 

dt (Bigec) == Oris Pi Cae i ene re 
a—(b+c)=a—b—c 
a—(b—c)=a—b-+e | 


(+a) xX (+6) =+ ab 
Ch a) b) rei) aa 
(—a)x (4+ 6)=—ab 
(— a) X (— 6) =+ ab 


The commutative law: 
Addition atb=b+a (3) 
Multiplication ab=ba | 
The associative law: 


Addition a+ (b -|- Cc) = (a ie b) a iy | é 
Multiplication a(bo) = (abe abe Wb 


INTRODUCTION. 25 


The distributive law: 
Multiplication a(b+c)=ab-+ae 


Enero Cl ol EN (5) 
ah a a 


The index law: 


Mirlirp li Ca Gon Meno en Grey tyr eee CG) 


These laws are true for all values of the letters, but in (6) 
m and n are for the present restricted to posztwe integral 
values. 


48, Value of an Algebraic Expression. Every algebraic ex- 
pression stands for a number; and this number, obtained 
by putting for the several letters involved the numbers for 
which they stand, and performing the operations indicated 
by the signs, is called the value of the expression. 

In finding the values of algebraic expressions, the begin- 
ner must be careful to observe what operations are actually 
indicated. Thus, 


4ameansa+a+ta-+ta; that is, 4 xa. 
a*meansaXaXaX a. | | 
Vabe means the square root of the product of a, b, and e. 


Vabe means the product of the square root of a by be. 


Nore. The radical sign V before a product, without a vinculum 
or a parenthesis, affects only the symbol immediately following it. 


Va-+ 6b means that b is to be added to the square root of a. 


a+ 6 means that b is to be added to a and the square 
root of the sum taken. 


49, In finding the value of a compound expression the 
operations indicated for each term must be performed before 
the operation indicated by the sign prefixed to the term. 


26 SCHOOL ALGEBRA. 


Indicated divisions should be written in the fractional form, 
and the sign X omitted between a figure and a letter, or 
between two letters, in accordance with algebraic usage. 
Thus, (6 — ¢)-+- 2 X e+ 26 should be written 


“= "426. 





Notz. The line between the numerator and denominator of the 
fractions serves for a vinculum, and renders the parenthesis un- 
necessary. 


If 6=4, and c= —4, the numerical value is 
goa) og Le ee on een eng mo! 
ax (--4) a at a 


Exercise 2. 


Note. When there is no sign expressed between single symbols 
or between compound expressions, it must be remembered that the 
sign understood is the sign of multiplication. 


If a=1, b=2, and ¢c=8, find the value of 


1. Ta— be. 5. 2a—b+e. 9. ~/4abe. 
2. act, ca ab + be — ae. 10. V6abe. 
3. 4ab—e. 7 P+ate’. 11. &— 8B. 
4. Gab—b—ec. 8. 2ab—5dc’. 12.) Ve —ia'. 
13. a—2(b+0). 16. 6b —10bc+ 12a42e. 
14. (a+ 6)?+2(c—a)*. 1%. 5 ey ee 
15. V6be—(b—c). 18. A/6 Bc: 


Ifa = 1,50 = 2, ‘c= Banc — 0. find the value of 
19. Ta—be+ 6d. 21. 4ab—cd—d. 
20. actb—d. 22. 2a—bte. 


INTRODUCTION. 27 


23. ab+ be—ad. 27. b—c+d. 

24. 2ab—5be. 28. a—2(b+0). 

25. V4abed. 29. 25(3—5c)+(a—2e). 
26. W4abcd. 30. 2(a+ 5). 


Exercise 3. 


Remove the parentheses (§ 20), and find the algebraic 


sum of 


ih, (Ge ae Gea TOS te (ie 6). 
(62) 3 (2B). 194 (2 Say 2 5. 
ait ee the Ty i cal anes ay gr 
De lt yaar aaa Vaca Liege Gey) 
ia Oe ee ety TEU 10 eco een 
spp Wes aeseea) Le een (Ss eee La 
7h Tay Gey g 1 ee Bes eT 
8. 2—(8—38-+ 4). 18. —(2—3-+44)—1. 
De (De Oy-1 (82) 19.) (10 8-12 110), 
10. —7+(8+2-—4). 20. +(8—2—3-—l). 


If a=1, d=2, and c= — 8, find the value of 
21. a+tb-+e. 24. a -(—)b)+e. 
22. a—b+e. 25. a—(—)b)—e. 
23. a—b—ce. 26. (—a)+(—b)4+ (—e). 


CEA PTR ci, 


ADDITION AND SUBTRACTION. 
INTEGRAL EXPRESSIONS. 


50. If an algebraic expression contains only zntegral 
forms, that is, contains no letter in the denominator of 
any of its terms, it is called an integral expression. Thus, 
e+Tev’—c—5e’x, Lax—tbcy, are integral expressions, 
hue aa 4b—c—x 

v7—ab+ 0 

An integral expression may have for some values of the 
letters a fractional value, and a fractional expression an 
integral value. If, for instance, a stands for $ and 6 for 


4, the integral expression 2a¢— 506 stands for $¢— 4; 


is a fractional expression. 


eee 
, he 
and the fractional expression ie stands for 15+2=5. 


Integral and fractional expressions, therefore, are so named 
on account of the form of the expressions, and with no refer- 
ence whatever to the numerical value of the expressions 
when definite numbers are put in place of the letters. 


51, A term may consist of a single symbol, as a, or may 
be the product of two or more factors, as 6a, ad, 5a’?be. If 
one of the factors is an arithmetical symbol, as the factor 5 
in 5a’bc, this factor is usually written first, and is called 
the coefficient of the term; the other factors are called literal 
factors. 

Note. By way of distinction, a factor expressed by an arithmeti- 


‘cal figure is called a numerical factor, and a factor expressed by a 
letter is called a literal factor. 


ADDITION AND SUBTRACTION. 29 


52. Like Terms. Terms which have the same combina- 
tion of literal factors are called like or similar terms; terms 
which do not have the same combination of literal factors 
are called unlike or dissimilar terms. Thus, 5a7bc, —7a’bc, 
a’be, are like terms, but 5a’be, 5ab’c, 5abc’, are unlike 
terms, 


53, A simple expression, that is, an expression of one 
term, is called a monomial. A compound expression, that 
ig, an expression which contains two or more terms, is 
called a polynomial. A polynomial which contains two 
terms is called a binomial, and a polynomial which contains 
three terms is called a trinomial. 


54. A polynomial is said to be arranged according to 
the powers of some letter when the exponents of that letter 
either descend or ascend in the order of magnitude. Thus, 
3axz*>— 4b2?— 6ax-+ 8b is arranged according to the de- 
scending powers of x, and 8b—6axr—462?+ 8az’* is 
arranged according to the ascending powers of wz. 


55. Addition of Integral Expressions. The addition of two 
algebraic expressions can be represented by connecting the 
second expression with ‘the first by the sign +. If there 
are no like terms in the two expressions, the operation is 
algebraically complete when the two expressions are thus 
connected. 

If, for example, it is required to add m+n—~p to 
a-+6b-+ce, the result will be a+d+e+(m+n—~p); or, 
removing the parenthesis (§ 20), a+0+ce+m+n—p. 


56. If, however, there are like terms in the expressions to 
be added, the like terms can be collected; that is, every 
set of like terms can be replaced by a single term with a 
coefficient equal to the algebraic sum of the coefficients of 
the ike terms. 


30 SCHOOL ALGEBRA. 


_ If it is required to add 5a?+4a+83 to 2a?—3a—4, 
the result will be 


20 —3a—4++ (507+ 4a+838) 


=2a?—8a—4+50?+4a+38 § 20 
=2¢+5a’—3a+4a—4+38 § 26 
=Tav’+a-—l. 


This process is more conveniently represented by arrang- 
ing the terms in columns, so that like terms shall stand in 
the same column, as follows: 


20 —8a—4 
5¢7+4a+8 
Yv+ a-—l 


The coefficient of a? in the result will be 5+ 2, or 7; the 
coeficient of a will be —3+4, or 1; and the last term is 
—4+3, or —1. 

Nore. When the coefficient of a term is 1, it is not written, but 


understood ; conversely, when the coefficient of a term is not writ- 
ten, 1 is understood for its coefficient. 


If we are to find the sum of 2a*— 3a7b+ 4ab?+ 6° 
a+ 4a7b — Tab? — 26°, — 8a? + ab — 8ab? — 46°, and 
20° + 2076 + 6ab?— 306°, we write them in columns, as 


follows : 20—80b+4ae?+ 6 
a’ + 4076 — Tab? — 28° 

—3a*+ ab—3ab’?— 45° 

2a? +2076 + 6ab? — 36° 


2a’? + 407) — 86° 


The coefficient of a* in the result will be 2+1—3-442, 
or +2; the coefficient of ab will be —8+4+1-+2, or +4; 
the coefficient of ab? will be 4—7—3-+ 6, or 0; and the 
coefficient of 83 will be 1—2—4-—8, or —8. 


12. 


13. 


14. 


15. 


16. 


17. 


ADDITION AND SUBTRACTION. 31 


Exercise 4. 


Add 

Oy aCe, and — id. 
-Tay, 2xy, —4ay, and — hay. 
4076, —3a’%b, and — 5a’. 


3xy, 4ay, Tax, and — saz. 

a+6 and a—8. 

x’? —x and 2° — 2’, 

527+ 627—2 and 3827—72+2. 

382’°—2aey+y and 2 —2ay4+ 37’. 

ax + ba —4, 3 aa? — 2ba-+ 4, and — 4az? —2bx-+5. 
5a+3y+z2, 34+2y+ 382, and x—38y—5z. 


~ 8ab—2a27+ 8072, 4ab—b6a'x+5az’, ab+a’x—az’, 


and az’— 8ab — 5a’z. 


at’ — 2a + 38a?—a+7, 2at— 3a? +2e?—a+6, and 
—a—2a+2d¢—5. 

38a7—ab+ac— 30?'+4be— ee, —5a?—ab—ac+5be, 
—4b6e+ 5c?+ 2ab, and —4a?+ 0? —5be+ 2c’. 

at — 8a +20? —4e+7, 82t+2e+2?—5ae—6, 
—4a*+32°—382°+9x2—2, and 22*—234+2?—2+1. 

82y—4ary+2*, 52°—llay—1222, —Ty +a2’°y— 22, 
and — 4227+ 7°— 2’. 

ai — 20+ 8a, &+ae’+a, 4a'+5a', 2¢7°+3a—2, 
and —a’—2a—8. 


+227 —xy—y’, 20 —8x’y —4ay—Ty’, 


and 2° — 8 ay’ — 77. 


B2, SCHOOL ALGEBRA. 


57. Subtraction of Integral Expressions. The subtraction 
of one expression from another, if none of the terms are 
alike, can be represented only by connecting the subtra- 
hend with the minuend by means of the sign —. 

If, for example, it is required to subtract a+6-+¢ from 
m-+n—p, the result will be represented by 


m+tn—p—(atb+e); 
or, removing the parenthesis, § 20, 
mtn—p—a—b—e, 


If, however, some of the terms in the two expressions are 
alike, we can replace two like terms by a single term. 
Thus, suppose it is required to subtract a’—2a’+2a—1 
from 3a*— 2a°+a—2; the result may be expressed as 
follows : 
30° —2a¢+a—2— (a —2a?+2a—1); 
or, removing the parenthesis (§ 20), 
8a°— 270 +a—-2—a+2e0—2a+l1 
= 3a°— ai—2a'+ 20 +a—2a—2+4+1 
= 2a'—a—l. 


This process is more easily performed by writing the 
subtrahend below the minuend, mentally changing the sign 
of each term in the subtrahend, and adding the two expres- 
sions. Thus, the above example may be written 


8A0—2¢0V+ a—2 

a— 2A 2 2a—1 

Di ie — a—l 
The coefficient of a will be 8—1, or 2; the coefficient 
of a? will be —2-+-2, or 0, and therefore the term a? will 


not appear in the result; the coefficient of a will be 1 — 2, 
or —1; the last term will be —2+1, or —1. 


ADDITION AND SUBTRACTION. 85 


Again, suppose it is required to subtract a°+ 4 a°u’-— 3 a’2' 
—4az* from a’a’?+2a?2*—4ax‘. Here terms which are 
alike can be written in columns, as before : 


a'a? + 2a’ — 4azx* 
a’ + 4a3a? — 8 a’2* — 4az* 
—@ — 3a°x’? + 5a’2’ 
There is no term a in the minuend, hence the coefficient 
of a in the result is O—1, or -—1; the coefficient of a*z’? 
will be 1~4, or —8; the coefficient of a’z*® will be 2+38, 


or +95; the coefficient of az* will be —4+4, or 0, and 
therefore the term az* will not appear in the result. 


Fixercise 5. 

From 8a — 46 — 2c take 2a— 36 — 3c. 
Froms we 40 1-S atate Ya 8 ora a. 
From 7a?—92—1 take 5a7— 6x—3. 
From 22? — 2aa+@ take 2? — ax — a’. 
From 4a — 86 — 8¢ take 2a— 386 +4e. 
From 527+ 72+4 take 32?— 72+ 2. 
From 2az-+ 3by +5 take 3ax2 — 3 by — 5. 
From 4a?— 6a6+ 20? take 8¢ + ab-+ 6. 


From 4076 + 7ab?+ 9 take 8 — 3 a0”. 


CO MP TP T Pw YE 


10. From 5a’c+ 6a’) — 8a? take 6° + 6a7b — 5a’c. 

11. From a? — 0’ take 6’. 13. From 0? take a? — 0’. 
12. From a? — 0? take a’. 14. From a take a? — 0’. 
15. From z*+ 8a2°— 262? + 3cx—4d 


take 8a*+ az’?—40?+ 6cx+4+d. 


34 SCHOOL ALGEBRA. 


If A=8a?—206+458)) -C=]te— 8ab +50’, 
B=9¢0—5ab4+30', D=1ld—sab—40* 


find the expression for 


16. A+C+ B+ D. 19. A+C—B-—D. 
17. A—C-—B+D. 20. A—C+ B+ D. 
18. C-A—B+ D. 21. A+C—B+D. 


58. Parentheses. From the laws of parentheses (§ 20), we 
have the following equivalent expressions : ; 


at(b+tec)=atb+e «.a+b+e=a+(b+c); 
a+(b—c)=a+b—c¢, »«.a+b—c=a+(b—¢); 
a—(b+c)=a-—b—c, «.a—b—c=a—(b+0); 
a—(b—c)=a—b+e «..a—b+e=a—(b—e); 


that is, if a parenthesis is preceded by the sign +, the 
parenthesis may be removed without changing any of the 
signs of the terms within the parenthesis; conversely, any 
number of terms may be enclosed within a parenthesis 
preceded by the sign +, without changing the sign of any 
term. 

If a parenthesis is preceded by the sign —, the paren- 
thesis may be removed, provided the sign of every term 
within the parenthesis is changed, namely, + to — and — to 
++; conversely, any number of terms may be enclosed with- 
in a parenthesis preceded by the sign —, provided the sign 
of every term enclosed 1s changed. 


59, Expressions may occur with more than one paren- 
thesis. In such cases parentheses of different shapes are 
used, and the beginner when he meets with a ( or a[ ora 
} must look carefully for the other part, whatever may in- 
tervene,; and all that is included between the two parts of 


ADDITION AND SUBTRACTION. 35 


each parenthesis must be treated as the sign before it directs, 
without regard to other parentheses. It is best to remove 
each parenthesis in succession, beginning with the innermost 
one. Thus, 


(1) a—[b—(e—d) +e] 
=a—[b—c+d+e] 
=a—bte—d—e. 

(2) G10 le (d—e) +f |i 


=a-{o-[e—dtet/]} 
=a—{b—c+d—e—f} 
=a—b+ce—d+e4+f. 


Exercise 6. 


_ Simplify the following by removing the parentheses and 
collecting like terms: 


1. a—b—[a—(6—c)— Cc]. 

m —[n—(p—m)]. 

22 —jy + [42—(y + 22)]5. 

8a —{26—[5e—(8a+4)]}. 
a—{b+[e—(d—b)+a]— 28}. 
382—[9—(22+ 7)+ 32]. 

2a —[y—(#— 2y)]. 

a —[26+(8¢e—26)+a]. 
(a—az+y)—(b—2—y)+(a+b—2y). 

8a —[—46+ (4a —b)—(2a—56)]. 

. 4e—[a—(26 —38c)+c¢]+[a—(26—5c—a)]. 
- e+(y—2)—[(82—2y) + 2]+[¢—(y— 22)]. 
. a—[2a+(a—2a)+ 2a]—5a—{6a—[(a+2a)—a]}. 


a 
0 2m f° oO 


36 SCHOOL ALGEBRA. 


14. 27 —(8y +2) ~{b—(e—b)+e¢ —[a—(e—B)}]}. 
15. a—[6+ce—a—(a+b)—c]+(2a—b+ 0), 


Notre. The sign — which is written in the above problem before 
the first term 6 under the vinculum is really the sign of the vinculum, 
—b +c meaning the same as —(b + ¢). 


16. 10N—2x—{—2—[z—(«4—5—2z)}}. 

17. Qa—§2Qa+(y—z)—824[22—(y—z—2y)—82]+-4y}. 
18. a—[b—{—ce+a—(a—b)—c}]+[2a—(6—a)]. 
19. a@— jb =[a—(e—b) 4+ ca — (a —b —c)—a|+ a}. 
20. 5a—{—8a—[8a—(2a—a—b)—a]+a}. 





Exercise 7. 


In each of the following expressions enclose the last three 
terms in a parenthesis preceded by the sign —, remember- 
ing that the sign of each term enclosed must be changed. 


1. 2a—6—8c—d+8e-—5f. 
x2—a—y—b—z—e. 
atb—c+4a—6+1. 

ax + by + ez + bx — cy + cz. 
8a+2b6+2c—5d—3e—4f. 
x—y+z2—d2y —422+4 d3yz. 


Considering all the factors that precede a, y, and z, respectively as 
the coefficients of these letters, we may collect in parentheses the 
coefficients of x, y, and z in the following expression : 


ae. OU We ON ae BS 


ax — by + ay —az—cz + bx =(a + b)a + (a—b) y—(a+ c)z. 
In hke manner, collect the coefficients of x, y, and z in 
the following expressions: 
7 axt by+cz+ bx—cy + az. 
8. ax-+ 2ay + 4az— ba + 8y — 8 bz — 2z. 


ADDITION AND SUBTRACTION. 387 


ax — 2by —5ez—4ba+ 38cy — Taz. 


: ax + Bay + 2by — bz—1lex + cy — cz. 

. 4by —3axe —6cz + 2bx — Tex — Bey — cx — cy — cz. 
- 6az— 5by + 8cz — 2bz — 8 ay + bz — ax + by. 

. 2—by+3az—8cy+ 2ax—2me—5bz. © 


. 2+ ay—az—acx + bez —-mny —y —z. 


Exercise 8. 
EXAMPLES FOR REVIEW. 
Add 42? — 5a?— 5a2z’?+ 6a’2, 6a? +3a°+4a2?+ 2.072, 
19 az?—112?—15a’2, and 102°+-7 avev+5a?—18 az’. 


Add 8a6+3a+66, —ab+2a+4b, Tab —4a—88, 
and 6a +126 —2ab. 


Nors. Similar compound expressions are added in precisely the 


same way as simple expressions, by finding the sum of their coefti- 
cients. Thus, 3(%—y) + 5(a—y) —2(«—y) =6 (x —y). 


3. 
4. 


Add 4(5 — 2x), 6(5 — x), 3(5— 2x), and —2(5— 2). 

Add (a+ 6)2+ (6+c)¥Y+(ate)?, 6+.e)2 
+(ate)y+(at+b)2’, and (a+c)2’+(a+b)¥ 
+ (6+ c)2. 

Add (a+ 6)a+(b+e)y+(e+a)z, (6+e¢)2+(e+a)z 
-—(a+b)y, and (a+c)y+(a+6)z2—(6+¢e)z. 

From a’ — 2” take a? + 2axz + 2”. 

From 8a?+ 2axr+ 2 take a? — ax — 2’. 

From 827— 8ax+5 take 527+ 2axr+5. 

From a?+ 3 6?c + ab? — abe take ab? — abe + 0°. 

From (a+ b)2+(a+c)y take (a—b)%—(a—c)y. 


. Simplify 7a—{8a—[4a—(5a—2a)]}. 


38 SCHOOL ALGEBRA. 


12. Simplify 83a—jfa+6—[a+6+c—(a+b+c+4d)}]}. 

13. Bracket the coefficients, and arrange according to the 
descending powers of x 
vw? —ax—c'x’— ba + ba? — cz? + aa? — 2? — cx. 

14. Simplify v—(0?—c’)—[0?—(?—a’)|+[?—-(’—a’)]. 

16.0 @==1b =3) c= 0, And dee tor ehea val aero. 
a — 2b—{8e—d—[8a—(5b—ce—8d)|—208}. 

16. From 2d+1la+106—5e take 2c+5a—806. Find 
the value of each of these expressions when a, 8, e, 
and d have the values l, 3, 5, 7, respectively, and 
show that the difference of these values is equal to 
the value of their difference.. 

17. If a=1, b5=—3, e=— 5, d=0, find the value of 
vt 20?+3¢e+ 4d’. 

If a=3, b6=4, c=9, and 2s=a-+6-+c¢, find the 
value of 

18. s(s —a)(s— b)(s—c). 

19. s*+(s—a)’?+(s—6b)’?+(s—e). 

20. s*—(s— a)(s —b) —(s— 5) (s — ec) —(s —¢)(s — a). 

21. Ift=a+2b—38ce, y=b+2c— 38a, and z=c+2a 
— 306, show that x+y+2=0. 

22. Ift=a—264+3c, y=b—2c+3a, and z=c—2a 
+306, show that «+ y+2=2a+26+4 2e. 

23. What must be added to 2+ 5y’+ 32 in order that 
the sum may be 27 — 2’? 

24. What must be added to 5a3— 7a?b+ 3ab? in order 
that the sum may be a* — 2a7b — 2ab’ + 6? 

25. If H=5a°+3a@)b—-20', F=38a' -—7Ta’b — 8’, 

G=2e¢7b—A—-—8, H=avb—2e— 36 
find the expression for #—[/’— (@— #)]. 


CHAPTER III. 
MULTIPLICATION. 
INTEGRAL EXPRESSIONS. 


60. The 1aws which govern the operation of multiplica- 


tion are formulated as follows: § 47 
Go pees as oes The commutative law, 
aX (bc) =(ab)Xe¢=abe . . The associative law. 


b = ab 
a(b+c)=a ied k The distributive law. 


a (b—c) = ab — ae 


am xX a® = Ca eee ieee, 62. bn index aww. 
aX (+6) =-+ ab 
a xX (— 4) =—ab 


The law of signs. 
(—a) xb =—ab 


(—a) x (—b6)=+ ab 
61. Multiplication of Monomials. When the factors are 
single letters, the product is represented by simply writing 


the letters without any sign between them. Thus, the prod- 
uct of a, b, and c is expressed by abe. 


62. The product of 4a, 50, and 8c is 
4ax5bx38ce=4x5~x 8abe= 60abe. 


Note. We cannot write 453 for 4x53 because another mean- 
ing has been assigned in Arithmetic to 453, namely, 400 + 50 + 3. 
Hence, between arithmetical factors the sign < must be written. 


40 SCHOOL ALGEBRA. 


63. The product of a7b and a*0? is 


Cox ee Svebl aro a0. 


64. To multiply one monomial by another, therefore, 


Find the product of the coefficients, and to this product 
annex the letters, gwing to each letter in the product an index 
equal to the sum of rts indices in the factors. 


Nots. The beginner should determine first the sign of the product 
by the law of signs, and write it down; secondly, after the sign he 
should write the product of the coefficients; and lastly, each letter 
with an index equal to the sum of its indices in the factors. 


65. We may have an index affecting an expression as 
well as an index of a single letter. Thus, (adc)? means 
abe X abe, which equals aabbce, or a’b’c?. In like manner, 
(aocje— o°b°c > Thatas, 


The nth power of the product of several factors is equal to 
the product of the nth powers of the factors. 


66. By the law of signs, we have 
(— a) x (— 6) =+ ab, 


and (+ ab) x (—c) =— abe, 
that is, (— a) x (— 4) X (ec) = — abe; 
and (— abc) xX (— ad) = + abed, 


that is, . (—a) xX (—}4) xX (—¢) X (-d) =+ abed. 
It is obvious, therefore, that 


The product of an even number of negative factors will 
be positive, and the product of an odd number of negative 
factors will be negative. 


MULTIPLICATION. 4] 


67. Polynomials by Monomials. We have (§ 47), 
a(b-+c¢c)=ab-+ae. 
In like manner, 
a (b—c+d—e)=ab—ac+ ad—ae. 
To multiply a polynomial by a monomial, therefore, 


Multiply each term of the polynomial by the monomaal, 
and add the partial products. 


Exercise 9. 


Find the product of 


To ic and 5D; 6. Tab and 38ae. 

2. 3x and 8y. 7. —2aand Ta*x’y. 

3. 3a’ and 6a’. 8. —38a’b and — 8ab’. 

4. 3a and 2a’. 9. —5mnp’ and —4m’?nip’. 
5. 2mn and 3m’n. 10. — 8a’, 9% and —3 ab. 


11. —227y, v*y*, and —32’y. 

12. —3a*y, —2a07b, and — 2’y'a°D?. 
138. 5a+386 and 2a’. 

14. ab—be and 5arbe. 

15. ab—ac— be and abc. 

16. 6a5b — 7a7b’c and a’b’c. 

17. a2 +6?'—¢ and abe’. 

18. 5a?—30?+ 2c’ and 4ab%c’. 
19. abe— 8a°%bc’ and — 2ab’e. 

20. — xyz? + v?y*z and — 2 yz. 


21. —2m’np* — mnp’ and — m’np. 


42 SCHOOL ALGEBRA. 


5 yD 


22. x—y—zand — 382°y'2’. 
23. —327 and 2’? +27 —2. 
24. 38x—2y—A4 and 52’, 


68, Polynomials by Polynomials. If we have m+n-+>p 
to be multiplied by a+ 6+, we may substitute Jf for the 
multiplicand m+n-+p ($12). Then 


(a+6+¢)M=aM+bM+ cM. § 28 


If now we substitute for AZ its value m-+n- p, we shall 
have 
a(m+tn+p)+6(m+n+p)+e(m+n+p) 
=am-+an+ap+bm- bn-+ bp+em-+en+ep. 
That is, to find the product of two polynomials, 
Multiply every term of the multiplicand by each term of 
the multiplier, and add the partial products. 


69, In multiplying polynomials, it is a convenient ar- 
rangement to write the multipher under the multiplicand, 
and place like terms of the partial products in columns. 


(1) 5a — 6 b 
Sa AD 
15a?—18ab 

— 20 ab + 246? 


15a? — 38ab + 2406? 


We multiply 5a, the first term of the multiplicand, by 
3a, the first term of the multiplier, and obtain 15a’; then 
— 66, the second term of the multiplicand, by 3a, and ob- 
tain —18ab. The first line of partial products is 15a? 
—18a. In multiplying by —40, we obtain for a second 
line of partial products —20ab+ 246’, which is put one 
place to the right, so that the like terms —18ad and 


MULTIPLICATION. 43 


—20ab may stand in the same column. We then add the 
coefficients of the like terms, and obtain the complete prod- 
uct in its simplest form. 


(2) Multiply 474+ 38+ 52?— 62° by 4—62?—5z. 


Arrange both multiplicand and multiplier according to 
the ascending powers of z. 


8+ 42+ 52°— 62° 
4— 5a4— 62° 


12+ 162+ 2027 — 242° 
— 15x — 202? — 252° + 302* 
— 182? — 242° — 802* + 862° 


12+ 2«£—182?— 732 + 362° 





(3) Multiply 14+ 2a-+ 2*— 382? by 2 —2—2z2. 
Arrange according to the descending powers of 2. 


z*—38a?+224+1 


o— 247 —F 
v—8e1+2e*+ 2 
— 22° +62°—42°—24 
— 22+ +62°7—4x4—2 
xz’ — 52° + 722+ 227—62r—2 


(4) Multiply a? + 0? + c? — ab — be—ae by a+b+e. 
Arrange according to descending powers of a. 


@—ab—act Bb~— bet ¢ 


at 6+ ¢ 
&—ab—ae+ab?— abct+ac 
+ a’b —ab?— abe + 6 — Be + be? 
+a’e — abe—ac’ + he—b?+dé 


a — 8abe + 0° +é 


44 SCHOOL ALGEBRA. 


Norr. The student should observe that, with a view to bringing 
like terms of the partial products in columns, the terms of the multi- 
plicand and multiplier are arranged in the same order. 


70. A term that is the product of three letters is said to 
be of three dimensions, or of the third degree. In general, 
a term that is the product of m letters is said to be of n 
dimensions, or of the nth degree. Thus, 5abe is of three 
dimensions, or of the third degree; 2.a7b’c’, that is, 2aabdbce, 
is of six dimensions, or of the sixth degree. 


Tl. The degree of a compound algebraic expression is the 
degree of that term of the expression which is of heghest 
dymensions. 


72. When all the terms of a compound expression are of 
the same degree, the expression is said to be homogeneous, 
Thus, 2+ 382°y + 32y?+y7* is a homogeneous expression, 
every term being of the third degree. 


13. The product of two homogeneous expressions 1s homo- 
geneous. For the different terms of the product are found 
by multiplying every term of the multiplicand by each term 
of the multiplier; and the number of dimensions of each 
partial product is the sum of the number of dimensions of 
a term of the multiplicand and of a term of the multiplier 
counted together. Thus, in multiplying a’+ 6’+¢?—ab 
_—be—ae by a+b-+c, Example (4), each term of the mul- 
tiplicand is of two dimensions, and each term of the multi- 
plier is of one dimension; we therefore have each term of 
the product of 2-+-1, that is, three dimensions. 

This fact affords an important test of the accuracy of the 
work of multiplication with respect to the literal factors ; 
for, if any term in the product is of a degree different from 
the degree of the other terms, there is an error in the work 
of finding that term. 


MULTIPLICATION. 45 


74. Any expression that is not homogeneous can be 
made so by introducing a letter, the value of which is 
unity. ‘Thus, in Example (3), the expressions can be writ- 
ten z*—3 a’2? + 2aPx+a* and a —2a’*x—2a*. The prod- 
uct will then be 2’ — 5a’ + Tata? + 2a°a? — 6a®x — 2a", 
which reduces to the product given in the example, by 
putting 1 for a, 


75. It often happens in algebraic investigations that 
there is one letter in an expression of more importance 
than the rest, and this is therefore called the leading letter. 
In such cases the degree of the expression is generally called 
by the degree of the leading letter. Thus, a?2?+-bx-+c 1s of 
the second degree in x. 


Exercise 10. 


Find the product of 


1. +10 and x+ 6. 12. 24—-3 and z+ 8. 

2. x—2and «—8. 13. e—T and 22—1. 

3. «—38 and xz+5. 14. m—nand 2m+1. 

4. x+3 and x—3. 15. m—aandm-ta. 

5. «—llandz—l. 16. 82+ 7 and 22 —8. 

6. —2+2and—x—8, 17. 54—2y and 52+ 2y. 
7. —x—2and x— 2. 18. 83a—4y and 24+ 3y. 
8. —2+4 and «— 4. 19. +7 and a’ — 7. 

9. —x+7 and 2+ 7. 20. 2x? + 37’ and wv? + 7’. 
10. x— 7 and z+ 7. 21. ¢t+y+zande—y-+z. 
11. x—8 and 2248. 22. a+2y—z and x—y+2z. 


23. 2 —ayt+y and a’+2y+y’. 
24. m’—mn+nandm+n. 


25. wW+tmn+n and m—n., 


46 


ee 08 OU ae EO og et 


SCHOOL ALGEBRA. 


26. a? — 38ab+ 0 and a’? — 8ab — 8’. 

27. @&—Tat+2 and a&—2a+8. 

28. 227—38xy+4y and 327+ 42y — dy’. 

29. v+tayty and 2 —xz— 2’. 

30. 7+Y7+2—a2y—xz—yzandx+y+z. 
31. 4a?—10ab-+ 25070? and 56+ 2a. 

32. v+4y andy’ + 42. 

33. vw +2ry4+8 and y’?+ 22y — 8. 

34, v7?+0?+1—ab—a—banda+1+48. 
35. 327— 27° + 52 and 82°4 2y7 — 32’. 

36. VY +y + 2ay—24-—2y—1 andew+y—l1. 
37. a®™+ 2a™-'— 38a"? —landa+l. 

38. a®— 4a™!+ 5a"? +a"? anda—1l. 

39. at! — 4a"+4 20" — a’ and 2a —a’? +a. 


40. 2*—y" and a*+y"™. 


Exercise 11. 
Simplify : 
(a+6+c)(a+6—c) —(2ab—’). 
(m+n)m—[(m—n)y—n(n—m)]. 
[ae —(a'— 6)(6 + e)| — b[6 —(a—c)]. 
(x —1)( — 2) —8x(e#+8)+2[(¢+ 2)(#+4+ 1) — 3] 
4(a—8b)(a + 36)—2(a— 66)? —2(a? +60), 
(a+y+ 2%—2(y+2—2)—y(a@+2—y)—2(a+y— 2). 


5[(a— 6) x—cy]—2[a(#— y)— ba] 
—[38ax—(5c— 2a)y]. 


eo wmnnrt noon fF WW WO & 


wow — S&S Fe He Se Se Se Se eS 
Comoe OO saat Oo OU eo ate be 


wow wo WwW 
oOo wo 


MULTIPLICATION. 47 


Exercise 12. 


EXAMPLES FOR REVIEW. 


Multiply — 
z’*—x—19 by #+22—3. 
1+22+2' by 1—2#'+22°— 382. 
22°4+2+32 by 22-8274 22’. 
oa°+5—4x by 8+ 62?—7z. 
a+a—y by #—y + xy. 
38-+727—5x by 827—6x—102°+4 
6°-+- 6ab*— 407) by 2a7b — ab’— 8a’. 
at+tax—b by 2+ 527 —4. 
a—metnetr by tert. 


x’ —(a+6)x%+ab by x—c. 


.P+aeytayty by y—x. 

- 42°+ 97’ — bay by 4274 9y'?+4+ Bry. 

. @—8ae+5 by 2’ +4. 

- t—xy+ty* by a+aryt+y. 
san a Ae ee by a — Bs 

. am —amy™ + y™ by a™ + y™. 

- a*—a*+a*—1 by a*+1. 

. @+P+?+2ab—ac—be by a+b+e. 

. Simplify (a — 26) (6—2a)—(a—36) (46 —a)+ 2a. 
. If a=0, 6=1, and c=— lI, find the value of 


(a—b)(a—c)+e¢(8a—b—c)+ 2ac—(a—c)+26. 


- [(Qa+y)P+ (@—2y)'][(82—2y)— @e—s8y)'], 
. &(b—c)—B (a—c) +e (a—b)—(a—b) (a—c) (6—¢). 
. (2a—b)+26 (a+b)—3a—(a—b)+(a+b) (a—d). 


CHAPTER IV. 
DIVISION. 


INTEGRAL EXPRESSIONS. 


76. The laws for division are expressed in symbols, as 
follows : 


+ab+(+a)=+6 
—ab-+-(+a)=—6 





Law of signs. 44 
ee rat | “a ae 
—ab+(—a)=+6 
EE Be sit . . . Distributive law. § 47,(5) 
a a a 


77. The dividend contains all the factors of the divisor 
and of the quotient, and therefore the quotient contains the 
factors of the dividend that are not found in the divisor. 


Thus, we a; at ge: an 31e. 
C a —4a 








78. If we have to divide a® by a’, a® by at, at by a, we 
write them as follows: 





AeS = C02 = 60 
a aa 
a’ aaaaaa iin een 
j= =aa =aw=a’”; 
a aaaa 

4 





DIVISION. 49 


79. To divide one monomial by another, ROSS we have 
the following rule : 


Divide the coefficient of the dividend by the coefficient of 
the dwisor (observing the law of signs), and subtract the 
mdex of any letter in the dwisor from the w dex of that letter 
m the diwidend. 


Thus, AU Day; at = Tab; Bee —9 x’ yz; 


8 qin! 1 yP- pala 
4n—1 a 1 } p—6, r—1 
3a SLOP RT 


— Aare 


Norte. Since “ = 1, and also by the rule above given, a 
a : a” 


=a, it follows that a®=1. Hence, any letter which by the rule 
would appear in the quotient with zero for an index, may be omitted 
without affecting the quotient. 


80. To divide a polynomial by a monomial, we have, by the 
distributive law, the following rule: 


Divide each term of the diwidend by the divisor, and add 
the partial quotients. 


8ab+4ac—6bad_8ab, 4ac_ bad 
poe 2a ay, Gee ce 


=46+2c—3d. 
9atb’x—120°%b2?—3a7z_ Yatb’e 120°b2? 307% 


8a°x 8a'x se 
=a) -.4ab7 — Ll. 


4n+1_ 3n 4n+1 3n 
62 4x be Oo 4a = 8 y2nt?2 __ 2 nth 


2, gin 29 gin 9 gin 1 


Notre. Here we have 4n+1—(2n—1)=4n4+1—2n4+1=2n+4 2, 
and 8n —(2n—1)=3n—2n+1=n+1, as indices of « in the first 
and last terms of the quotient respectively. 


50 


_ 


— tet 
ree fe =) 


SCHOOL ALGEBRA. 


Exercise 13. 


Divide 
8a° by a’. 
—422" by 62°. 
— 352° by 52%. 
—62" by — 8z. 
202'y? by — 4xy’. 
— 21 2* by — Tx*. 


28 a'b® by — Tab?. 
’ al 25 2°0? by — 526. 


— 35 p’q’ by 5pq’. 


. —167’s’ by — 47°, 


- 28m"n" by 4m'n”. 


— 24 m*n® by —4 mn’. 


13. 


14. 
15. 
16. 
Vee 
. —50a°S* by — 10a*. 
. 840 a"? by 14 27%. 

. —380a'x"2z by — 622. 


abz’ by abs. 
—2a’bx* by aba. 

4 aba? by —azx’. 
186'2" by —9b'2". 
256 xyz" by 8x*y’z. 


- a+ 22xy by 2. 
. &—2ab by a. 
. 42°— 82° by 22’. 
24. 


—62°— 24 by — 22. 


25. — 8a°— 16a” by — 8a’. 

26. 27a*— 36a’ by 9a’. 

27. —30a'+ 20a’ by — 10a*. 
28. —122%y* — 42°77" by — 42°77’. 


29. —32"z' — 62°2* by —3.2°2*. 

30. 3a°b'c’ — Davbic’ by 8.abic’. 

31. 2? — xy — xz by —x. 

32. 38a’ -- 6a07b — 9ab* by — 38a. 

33. vy? — vy — vy by 2’. 

34. a®b’c — a’b®c — a’be® by abe. 

35. 8a°—4a*b — bal” by — 2a. 

36. 5m'n — 10m'n? — 15 mn’ by 5mn, 


DIVISION. 51 


81. To divide one polynomial by another. 


If the divisor (one factor) ree atb-+te, 
and the quotient (other factor)= n+p+q, 
an+bn-+en 
then the dividend (product) = 4 -+ap+obp+ep 
+aq+ bq+cq. 


The first term of the dividend is an; that is, the product 
ol a, the first term of the divisor, by x, the first term of the 
quotient. The first term n of the quotient is therefore 
found by dividing an, the first term of the dividend, by a, 
the first term of the divisor. 

If the partial product formed by multiplying the entire 
divisor by ” be subtracted from the dividend, the first term 
of the remainder ap is the product of a, the first term of 
the divisor, by p, the second term of the quotient; that is, 
the second term of the quotient is obtained by dividing the 
first term of the remainder by the first term of the divisor. 
In like manner, the third term of the quotient is obtained 
by dividing the first term of the new remainder by the first 
term of the divisor; and so on. Hence we have the fol- 
lowing rule: 


Arrange both the diidend and divisor mn ascending or 
descending powers of some common letter. 
“Dwide the first term of the dinidend by the first term of 
the disor. 
Write the result as the first term of the quotient. 
Multiply all the terms of the disor by the first term of 
the quotrent. 
Subtract the product from the diidend. 
Tf there be a remainder, consider it as a new dividend 
and proceed as before. 


52, SCHOOL ALGEBRA. 


82. It is of fundamental importance to arrange the divi- 
dend and divisor in the same order with respect to a com- 
mon letter, and to keep this order throughout the operation. 

The beginner should study carefully the processes in the 
following examples: 


(1) Divide 27+ 182+ 77 by «+7. 


SER eae 

e+ Tx z+1l 
lla+177 
llz+77 


Norr. The student will notice that by this process we have in 
effect separated the dividend into two parts, 22+ 7a and lla + 77, 
and divided each part by «+7, and that the complete quotient is 
the sum of the partial quotients x and 11. Thus, 

w+ 18ea4+77=09+ Te 4+ 1le+77=(2?4+ 72) +(1l2+4+ 77); 
Pes 477 at Te ae TT 


= a } = $11. 
x+7 e2+7 u+7 


(2) Divide a@—2ab+0 by a—O. 
@—2ab+6'la—b 
a@— ab a—b 
— abt 
— abt 


(3) Divide 4a‘x’?— 4a’2* + a6—a® by 2?—a?. 
Arrange according to descending powers of z. 
vw—4ect+4ate?—afl2’— ov 
x— a'r xz'— 8a’2’?+ a+ 
— 8a7x* + 4a‘z’? — a® 
— 8a7a*+ 8a‘2’ 


atx? — a® 
atz? — a? 


DIVISION. 53 


(4) Divide 22a7b? + 15d* + 3a* — 10a*) — 228° 
by a’ + 30’ — 2a6. 
Arrange according to descending powers of a. 


Zat — 10 a°d + 22.072? — 22.ab° +153] a?—2ab+38 
8at'— b6ad+ Yad? 3a? —4ab +56 
— 4% + 13076? — 22ad° 
— 40°64 8a7b?— 12ab' 
5a?b? — 10 ab? + 1584 
5a’b? — 10 ab®+ 150+ 






(5) Divide 528—2+1—32* by 1+ 32? —2z. 
Arrange according to ascending powers of 2. 
l— #+52°—82t|1—22+ 32° 
1— 22482? eae 
x — 32+ 5a’ — 324 
x— 297+ 82° 
— #4 22° — 32 
— 2#+927*— 32" 


(6) Divide 2° + 7° + 2— 82yz by x+y+z. 
Arrange according to descending powers of z. 


ee 
a yx" + 2x" v—yrx—waty—yte? 


— ye — 22° —38yzr+yYte2 


—yL—Yu— yr 
—2 + ya —2yr+ty te 
— 2x" = Yer — 22 
yxe— yeterrty+2 
yea +ytyz 
— yeeter—y2+e 
— YU —Y2—Yy2 
ga + yz + 2° 


Za + yz? + 2° 


54 


SCHOOL ALGEBRA. 


(7) Divide 4a7*’ ~ 80a” + 19a*" + 5a*” + 9a? 
by a*°— Ta®** + 2a**> — 8u" ©, 


4a*130a7+19 a+ 5a**4+9a"* 
Aatt_28 47+ 8971-120? 






eT a 2a" *— 3a" * 
4a*—2e°—30’ 


— 2at%+1la*1+17 a*“749 a? 
a 2a7+14 q?—1— 4a? 16 az? 


=e FH 41121 a?" 6a" 219 qz—* 
— Smt a bare to 


Nors. We find the index of a in the first term of the quotient by 
subtracting the index of a in the first term of the divisor from the 
index of a in the first term of the dividend. Now (x + 1)—(«#—83) 
=x+1l1—x2+3=4. Hence 4 is the index of a in the first term of 


the quotient, 


Divide 


CUT 14 09 80 


bie 
12. 
gL. 
14. 
15. 
16. 
17. 
18. 


a+ Ta+12 by oe ee Sent 
a@—d5a+6bya—3. 7%. 62°—11l2+4 by 3x—4, 
v+2xry+ybyz+y. 8 

4 


e—2xry+y by x—y. 
v—Yy by x—y. 10. a’— 8a—3 by 3—a. 


In the same way the other indices are found. 
Exercise 14. 


- 42°+122+4+9 by 22-43. 


- 82’—l0ax—3a’ by 4x44. 
. d3a’— 4a—4 by 2—a. 


a‘+1la@’—12a—5a’+6 by 8+ a’—38a. 
oy BY shy — ly se byagee a ae 
36 + m*— 138m by 6+ m?+ 5m. 
1—s—38s?—s° by 1428+ 8". 

b§— 26°+1 by &—26+1. 

w+ 2x +9y4 by 2?—2ry + 37. 

a’ + 6° by at—a*b + ab? — ab? + Ot. 
1+52°— 62* by 1—24+ 382". 


DIVISION. 55 


. 82777+ 9y*+ 162* by 427+ 38y — 4 zy. 

» P+Y¥+t74 3ey4+ 3x7 by trt+y+z. 

. &@+0+e—8abe by atb+e. 

. @487+2—6ayz by w+ 474 2—a22—Qary—Zyz. 
. 22°— 3y'-+ ry —az —4y2-—-2 by 22+3y-4 2. 

. &—y—Qyz—2 by x+y+2. 

~ e+e +y by #+ayt+y’. 

. @—9e’?+ 1224-4 by 2?+ 32 —2. 

. p—2ay— by +4yY+138y+6 by ¥+38y74+3y+1. 
- ¥ —d5Y27+42 by Y—yz— 22. 

. @—4y— 924 12y2 by 44+ 2y— 8z. 

. 2& —41¢4—120 by a? + 42+ 5. 

. &—38+5xex+e7—42 by 8-22-27". 

. 6— 22*+4+ 10a°—1l2z’+z2 by 42-3 — 22’. 

. 1—6a'+ 52° by 1— 224 2”. 

. @+814 92" by 382 —2?—9. 

. &—y by @+ay+y. 

. &+y° by a’+y’. 

- @&+aev’+tat by 2—-ar+a’. 

. @&—20?+ab—3c’+ Tbe+2ac by 8e+a—b. 

- ab+2a°— 3b’—4be—ac—c’ by c+ 2a+4 36. 

- lda*+10a'r + 4072? + 6az*— 82 by 8a?+ Zax — 2’. 
. &—8—1—6ab by a— 26-1. 

- @— Barry" + Bary™—y™ by a*—y". 


j qgmtngn be ad 4q™t" —1},2n we OT qmtn -253n + 42 qmtn—3hin 


by a™+ 3a" 6" —Ga™*b™. 


56 SCHOOL ALGEBRA. 


83. Integral expressions may have fractional coefficients, 
since an algebraic expression is integral if it has no letter in 
the denominator. ‘The processes with fractional coefficients 
are precisely the same as with integral coefficients, as will 
be seen by the following examples worked out: 


(1) Add 4a’—4ab+40’, and ga’+ gab— $30’. 
7 t¢@—tab+10° 
ba+2ab— 30 
ja + 4ab 487 


(2) From $a*— fab + 40° take 4a@°—4ab+ 80" 


Pe 172 
ta—tab+ 16 
bys asian, 5 }2 
ta?—tab+ 86 
Trees meek 2 


(3) Multiply $a7—3ab+10’ by 4a—320. 
1@7—tab + 10 
: ta —2b 
ta’—idb+ tab’ 
—4@b+4+ 2#ab?—10° 


1282 12 25e7o aes 
ta’—tab+73ab’—16 





(4) Divide 30°-+ 4Lbd?—4480'd—%-d* by $b—$. 


208 4485] + 11bd?— 6 d3|8b —Sd 

yo Hed cee: 
— 2 bd + ibd? — 5d 
— 2Bd+ bd? 


3 bd? —5,d3 
8bd?— Ad? 


a on F WO ww 


Cee Oe 


DIVISION. 57 


Exercise 15. 
Add 307 + 40?ct+ 2 and — 3,ab—10° a’ 
From $2°+ 3ax— $a’ take 22°—3ax —1a’. 
From 4y—3a—%2+46 take ty+4a— 2x. 
Multiply 4c?—}e—4 by 4¢°—4e4+Hh. 
Multiply $2 —42?+42° by $24+42°+ 12°. 
Multiply 0.5 m*—0.4m'n + 1.2m’n? + 0.8mn? — 1.4n* 

by 0.4m? — 0.6mn — 0.8’. 

Divide a‘—{fa’b+420°0’+1ab® by 3a+40. 
Divide —4d°+ d’?’— $4d°+ 8d‘ by —#d’?+ 2d. 


Exercise 16. 


EXAMPLES FOR REVIEW. 
Find the value of 2° + ¥°+ 2— 32yz, if x=1, y= 2, 
and z=—8. 


Find the value of V2be —a, and of V2bce—a, if b=8, 
€=9 and, @= 23. 


Add a’b — ab’+ 6° and a’ —40°b + ab’ — 30°. 
Multiply a”— a™b™+ b™ by a™-+ b”. 

Multiply 4a°*+ 6a™t*+ 9a? by 2a™t*— 3a’. 
Divide 2? + 87? — 1252+ 30ayz by w+ 2y—5z. 
Simplify (a2 — a)?—(«— b)?—(a— 6) (a+6—8382). 


Find the coefficient of x in the expression 


xta—2[2a— b(e—2)]. 
Moltiphy ae Fae er hae? by bat, 


58 


10. 
11. 
12. 
LER 
14. 
15. 
16. 


17. 
18. 


19. 


20. 


Zi. 


22. 


23. 
24. 
25. 
26. 
27. 


SCHOOL ALGEBRA. 


Dividetc’d? cde ec ee vance ge 

Divide my” — mity"*" + m>*y"* by m>*y" 4, 

Divide .a't*’ — a®+ a?t" by a?-*t”s. 

Divide 22 ash er veee iee 

Divide Y—y + yy?" by ym, 

Divide 24°" — 62"y" + 6a" — 2y™ by z*—y". 

Divide z°— 2aa2?+ aa — abe — bax +a@b+ ab’ 
by 2’?—ax-+ bx — ab. 

Divide o™+27"+ 1 by 2°™—a2™-+-1, 


Divide 8 a™t?— 4 y7t6— 129 y™t5 __ Q amt 
by at Bane, 

Divide 624715 — 18 a7™t* + 189°"15 — 184745 _ 55 
by 22°"— 382™—1. 


Divide 12a" — q**-? — 20a! + 19a" — 10a" 
by 4a" — 3a°t+ 2a’. 


Arrange according to descending powers of x the fol- 
lowing expression, and enclose the coefficient of each 
power in a parenthesis with a minus sign before each 
parenthesis except the first: 


ao — 2be — a? — ax — av? — cr — 2? — dcx. 


Divide 1.2a'x — 5.494 a°2? + 4.8072? + 0.9 az*— 2° 
by 0.6ax% — 22”. 


Multiply $a*—4ab + 30° by $a TO. 

Multiply $a?+ ab +30" by 4a— 40. 

Divide 4a°+A,ab?+ 7,0 by 4a+120. 

Subtract 42? + tay +1477 from 42?—4tay+ ty’. 
Subtract v’?-+ 42y —47 from 22? — tay+ y’. 


CHAPTER V. 
SIMPLE EQUATIONS. 


84. Equations. An equation is a statement in symbols 
that two expressions stand for the same number. Thus, 
the equation 82-+2=8 states that 32+ 2 and 8 stand for 
the same number. 


85. That part of the equation which precedes the sign 
of equality is called the first member, or left side, and that 
which follows the sign of equality is called the second mem- 
ber, or right side. 


86. The statement of equality between two algebraic 
expressions, if true for all values of the letters involved, 
is called an identical equation; but if true only for certain 
particular values of the letters involved, it is called an 
equation of condition. Thus, (a+6)’=a’?+2ab+ 0’, which 
is true for all values of a and 3b, is an identical equation ; 
and 82+2=8, which is true only when 2 stands for 2, is 
an equation of condition. 

For brevity, an identical equation is called an identity, 
and an equation of condition is called simply an equation. 


87. We often employ an equation to discover an unknown 
number from its relation to known numbers. We usually 
represent the unknown number by one of the das¢ letters of 
the alphabet, as x, y, 2; and, by way of distinction, we use 
the first letters, a, 6, c, etc., to represent numbers that are 
supposed to be known, though not expressed in the number- 


60 SCHOOL ALGEBRA. 


symbols of Arithmetic. Thus, in the equation az+b=ce, 
x is supposed to represent an unknown number, and a, 6, 
and ¢ are supposed to represent known numbers. 


88. Simple Equations. An integral equation which con- 
tains the first power of the symbol for the unknown number, 
x, and no higher power, is called a simple equation, or an 
equation of the first degree. Thus, av-+d—=—c is a simple 
equation, or an equation of the first degree wm x. 


89. Solution of an Equation. To solve an equation is to 
find the unknown number; that is, the number which, when 
substituted for its symbol in the given equation, renders the 
equation an identity. This number is said to satzsfy the 
equation, and is called the root of the equation. 


90. Axioms. In solving an equation, we make use of 
the following axioms: 


Ax. 1. If equal numbers be added to equal numbers 
the sums will be equal. 


Ax. 2. If equal numbers be subtracted from equal num- 
bers, the remainders will be equal. 


Ax. 8. If equal numbers be multiplied by equal numbers, 
the products will be equal. 


Ax. 4. If equal numbers be divided by equal numbers, 
the quotients will be equal. 


If, therefore, the two sides of an equation be increased by, 
diminished by, multiphed by, or divided by equal numbers, 
the results will be equal. 

Thus, if 8a = 24, then 84+4= 244-4, 8r—-4= 24 —4, 
4x 8a4=4~x 24, and 8a4+4= 24-4. 


91. Transposition of Terms. It becomes necessary in solv- 
ing an equation to bring all the terms that contain the 


SIMPLE EQUATIONS. 6l 


symbol for the unknown number to one side of the equation, 
and all the other terms to the other side. This is called 
transposing the terms. We will illustrate by examples: 


(1) Find the number for which z stands when 
162—11=7x-+ 70. 


The first object to be attained is to get all the terms 
which contain z on the left side of the equation, and all the 
other terms on the right side. This can be done by first 
subtracting 72 from both sides (Ax. 2), which gives 


9x —11= 70, 
and then adding 11 to these equals (Ax. 1), which gives 
92 = 81. 


If these equals be divided by 9, the coefficient of x, the 
quotients will be equal (Ax. 4); that is, x= 9. 
(2) Find the number for which x stands when x+b=a. 
The equation is z+b=a. 
Subtract 5 from each side, 7 +b—b=a—Bb., (Ax. 2) 
Since +6 and — 6 in the left side cancel each other 
(§ 36), we have z=a—b. 
(3) Find the number for which x stands when x—b=a. 
The equation is 2z—b=a. 
Add +6 to each side, x—b+b=a+b. (Ax. 1) 
Since —b and +46 in the left side cancel each other. 
(§ 36), we have z=atb. 
(4) What number does x stand for when ax+b=cz+d? 


This is the general form which every simple equation in 
x will assume when the like terms on each side have been 


/ 


62 SCHOOL ALGEBRA. 


collected. In this equation x represents the unknown num- 
ber, and a, 0, ec, d represent known numbers. 
If now we subtract (Ax. 2) cx and 6 from each side of 
the equation, we have 
ax —cx=a—b; 


or, bracketing the coefficients of z, 





(a—c)x ea SIA 

Whence, dividing both sides by a —e, the coefficient of z, 
we get d any, 
a ° 
a—ec 


92. The effect of the operation in the preceding equa- 
tions, when Axioms (1) and (2) are used, is to take a term 
from one side and to put it on the other. side with its sign 
changed. We can proceed in a like manner in any other 
case. Hence the general rule: 


93. Any term may be transposed from one side of an equa- 
tion to the other provided its sign 1s changed. 


94, Any term, therefore, which occurs on both sides with 
the same sign may be removed from both without affecting 
the equality. 


95. The sign of every term of an equation may be 
changed, for this is effected by multiplying by — 1, which 
by Ax. 3 does not destroy the equality. 


96. Verification. When the root is substituted for its 
symbol in the given equation, and the equation reduces to 
an ¢dentity, the root is said to be verified. We will illustrate 
with examples : 

(1) What number added to twice itself gives 24? 


Let x stand for the number ; 


SIMPLE EQUATIONS. 63 


then 2% will stand for twice the number, 
and the number added to twice itself will be x+ 2a. 
But the number added to twice itself is 24; 


Gh isa OS, 
Combining 2 and 2z2, 3x = 24. 
Divide by 3, the coefficient of z, e= 8. (Ax. 4) 
The required number is 8. 
VERIFICATION. c+ 2a = 24, 
8+2x 8= 24, 
8+ 16 = 24, 
24 = 24. 


(2) If 42—5 stands for 19, for what number does x 
stand ? 


We have the equation 4¢—5=19. 
Transpose — 5 to the right side, 42=—19+ 5. 
Combine, 4a = 24. 
Divide by 4, x= 6. (Ax. 4) 
VERIFICATION. 4%—5=19, 
4x6—5=19, 
24 -—5=19, 
19 = 19: 


(3) If 3¢—7 stands for the same number as 14 — 4z, 
what number does z stand for? ia he 


We have the equation 82—-T=14—A4z. 


Transpose 42x to the left side, and 
7 to the right side, 82+44%=—14+4 7. 
Combine, rt Wee ob. 


Divide by 7, esting: 


64 | SCHOOL ALGEBRA. 


VERIFICATION. 8x2—7=—14—4z2, 
8x3—T=14—4~x 3, 


2= 2. 
(4) Solve the equation (x — 8) (w— 4) =a (4 — 1) — 30. 
We have (a — 8) (w — 4) = 2 (x— 1) — 30. 


Remove the parentheses, 
v—Tea+12=2—-2— 30. 
Since z? on the left and 2 on the right are precisely the 
same, including the sign, they may be cancelled. 
Then —Te+ 12 =— x— 80. 
Transpose — x to the left side, and + 12 to the right side, 
—Tztx=— 380-12. 
Combine, 
— 6x =— 42. 
Divide by —6, z=. 
VERIFICATION. | 
(7 — 3) (7—4) = 7(7—1) — 30, 
4x3=T7~x 6—80, 
12 = 42 — 30, 
12= 12. 


Exercise 17. 


Find what number 2 stands for: 


1. If x—5D stands for 7. 6. If24—5=7+2. 
2. If x+8 stands for 12. 7. If227—4=5—z2. 
8. If 6z—12stands for 18. 8, IfS52—2=82+4. 
4. If 7~—8 stands for 25. 9. If7z—-5=62—1. 
5. If 52+8 stands for 48. 10. If 5a—3 = 25 — 2z. 


x 


SIMPLE EQUATIONS. | 65 


11. If 3x and x+ 8 stand for the same number. 


12. If 22—5 stands for the same number as By 


Solve the equations: 


13. 2x—8=8+ 42. 16. 8a2—4=12—2. 

14, 5a+4=2042. Vib Da Toe. 

15. 24—-3=7—z, 18. 82+14=2—z. 
Find x 


19. If 2x—5 and 42 —11 stand for the same number. 
20. If x(x—7) and x?-- 70 stand for the same number. 
21. If x(82—2) and 8x(¢a—1)+ 2 stand for the same 


number. 

- 22. If 8a—5=427—-10. . 

23. If 2a—4=—14—z2z. 

24. If 3u—8 and 4x —11 stand for the same number. 
25. If 2+—5 and 7—~2 stand for the same number. 

26. If 22?— 23 and (2%+1)(#—3) stand for the same 


number. 
27. If (+3) (@— 7) —(@—4) (# + 1) = 25. 
28. If (22—1)(«+8)—(e—8) (2u—8)= 72. 
Solve the equations: 
29. «(«—5)=2?— 80. 
30. (7+ 3) =27+ 18. 
31. («—3)(#4+1)=2—32+1. 
32. (x—1)(#+2)+(a+3) (#—1)=22(2+4)—(xt)). 
33. «(+ 8)—(@+1)(e@—2)—5(@#+3)4+3=0. 
34. (z—8)(@+38)—(w—4)@+4)—2=0. 


66 SCHOOL ALGEBRA. 


97. Statement and Solution of Problems. The difficulties 
which the beginner usually meets in stating problems will 
be quickly overcome if he will observe the following direc- 
tions: 

Study the problem until you clearly understand its mean- 
ing and just what is required to be found. 

Remember that x must not be put for money, length, 
time, weight, etc., but for the required number of specified 
units of money, length, time, weight, etc. 

Express each statement carefully in algebraic language, 
and write out in full just what each expression stands for. 

Do not attempt to form the equation until all the state- 
ments are made in symbols. 

We will illustrate by examples: 


(1) John has three times as many oranges as James, and 
they together have 32. How many has each? 


Let w be the number of oranges James has; 
then 3a is the number of oranges John has; 
and «+32 is the number of oranges they together have. 


But 32 is the number of oranges they together have; 


“ 2£+3u = 32; 

or, 4a = 32, 
and x = 8. 
Since 2 = 8, 3a = 24. 


Therefore James has 8 oranges, and John has 24 oranges. 
Nore. Beginners in stating the preceding problem generally write: 
Let « = what James had. 


Now, we know what James had. He had oranges, and we are to 
discover simply the nwmber of oranges he had. 


(2) James and John together have $24, and James has 
$8 more than John. How many dollars has each? 


SIMPLE EQUATIONS. 67 


Let x be the number of dollars John has; 
then x + 8 is the number of dollars James has; 
and a + (a + 8) is the number of dollars they both have. 

But 24 is the number of dollars they both have; 
“. @ +(x + 8) = 24. 
Removing the parenthesis, we have 
r+e+8 = 24. 
Transposing and collecting like terms, we have 
2% = 16. 
Dividing by 2, we get 
x= 8. 


Since az = 8, o+8 = 16, 
Therefore John has $8, and James has $16. 


Norse. The beginner must avoid the mistake of writing 
Let « = John’s money. 


We are required to find the number of dollars John has, and there- 
fore « must represent this required number. 


(3) The sum of two numbers is 18, and three times the 
greater number exceeds four times the less by 5. Find the 
numbers. 


Let «= the greater number. 
Then, since 18 is the sum, and z is one of the numbers, the other 
number must be the sum minus x Hence 


18 ~ x = the smaller number. 


Now, three times the greater number is 3, and four times the less 
number is 4(18 — 2). 

We know from the problem that 3a” exceeds 4(18 — x) by 5; or, 
in other words, we know that the excess of 3a over 4(18 — x) equals | 
5. It only remains to determine what sign the word “excess” im- 
plies. If we are in doubt about it, we can apply the phrase to two 
arithmetical numbers. We shall have no difficulty in seeing that the 
excess of 50 over 40 is 10; that is, 50~ 40, and hence that the sign 

— is implied by the word “ excess,” 


68 SCHOOL ALGEBRA. 


Hence, 3” —4(18 — x) = the excess 
But 5 = the excess. 
“. 3@—4(18 —2) =5, 
or 32—72+44%=5., 
“. 7e= 77, 
and v= 11. 


Therefore the numbers are 11 and 7. 


(4) Find a number whose treble exceeds 40 by as much 
as its double falls short of 35. 


Let x = the required number ; 
then 3a = its treble, 
and 3a — 40 = the excess of its treble over 40; 
also, 35 — 2a” = the number its double lacks of 35. 
Hence, 32—40 = 35 — 22. 
Transposing, 32 +22 = 35 + 40, 
“. 0% = 75, 
and go == 15, 


Therefore the number required is 15. 


(5) Find a number that exceeds 50 by 10 more than it 
falls short of 80. 


Let « = the required number ; 
then x — 50 = its excess over 50, 
and 80 — a= the number it lacks of 80. 
Hence, x — 50 — (80 — x) = the excess. 
But 10 = the excess. 
“. « — 50 — (80 — x) = 10, 
or 2—50—80+2=10. 
“. 22 = 140, 
and 2 = 70, 


Therefore the number required is 70. 


SIMPLE EQUATIONS. 69 


Exercise 18. 


1. If a number is multiplied by 7, the product is 301. 
Find the number. 


2. The sum of two numbers is 48, and the greater is five 
times the less. Find the numbers. 


3. The sum of two numbers is 25, and seven times the 
less exceeds three times the greater by 35. Find the num- 
bers. 


_ 4. Divide 20 in two parts such that four times the 
greater exceeds three times the less by 17. 


5. Divide 23 into two parts such that the sum of twice 
the greater part and three times the less part is 57, 


6. Divide 19 into two parts such that the greater part ex- 
ceeds twice the less part by 1 less than twice the less part. 


7. A tree 84 feet high was broken so that the part 
broken off was five times the length of the part left stand- 
ing. Required the length of each part. 


8. Four times the smaller of two numbers is three times 
the greater, and their sum is 68. Find the numbers. 


9. A farmer sold a sheep, a cow, and a horse for $216. 
He sold the cow for seven times as much as the sheep, and 
the horse for four times as much as the cow. How much 


did he get for each? 


10. Distribute $15 among Thomas, Richard, and Henry 
so that Thomas and Richard shall each have twice as much 


as Henry. 


11. Three men, A, B, and 0, pay $1000 taxes. B pays 
four times as much as A, and C pays as much as A and B 
together. How much does each pay? 


70 SCHOOL ALGEBRA. 


12. John’s age is three times the age of James, and 
their ages together are 16 years. What is the age of each? 


13. Twice a certain number increased by 8 is 40. Find 
the number. 


14. Three times a certain number is 46 more than the 
number itself. Find the number. 


15. One number is four times as large as another. If I 
take the smaller from 12 and the greater from 21, the 
remainders are equal. What are the numbers? 


16. The joint ages of a father and son are 70 years. If 
the age of the son were doubled, he would be 4 years 
younger than his father. What is the age of each? 


17. A man has 6 sons, each 4 years older than the next 
younger. ‘The eldest is three times as old as the youngest. 
What is the age of each? 


18. Add $24 to a certain amount, and the sum will be 
as much above $80 as the amount is below $80. What is 
the amount? 


_ 19. Thirty yards of cloth and 40 yards of silk together 
cost $330; and the silk costs twice as much per yard as’ 
the cloth. How much does each cost per yard? 


20. Find the number whose double diminished by 24 
exceeds 80 by as much as the number itself is less than 100. 


21. In a company of 180 persons composed of men, 
women, and children there are twice as many men as 
women, and three times as many women as children. 
How many are there of each? 


22. A banker was asked to pay $56 in five-dollar and 
two-dollar bills in such a manner as to pay the same num- 
ber of each kind of bills. How many bills of each kind 
must he pay? 


SIMPLE EQUATIONS. 71 


23. How can $3.60 be paid in quarters and ten-cent | 
pieces so as to pay twice as many ten-cent pieces as 
quarters? 


24. I have $1.98 in ten-cent pieces and three-cent pieces, 
and have four times as many three-cent pieces as ten-cent 
pieces. How many have I of each? 


Norr. In problems involving quantities of the same kind ex- 
pressed in different units, we must be careful to reduce all the quanti- 
ties to the same unit. 


25. I have $17 dollars in two-dollar bills and twenty- 
five-cent pieces, and have twice as many bills as coins, 
How many have I of each? 


26. I have $6.50 in silver dollars and ten-cent pieces, 
and I have 20 coins in all. How many have I of each? 


27. A bought 9 dozen oranges for $2.00. Fora part he 
paid 20 cents per dozen; for the remainder he paid 25 cents 
a dozen. How many dozen of each kind did he buy? 


28. A gentleman gave some children 10 cents apiece, 
and found that he had just 50 cents left. If he had had 
another half-dollar, he might have given each of them at 
first 20 cents instead of 10 cents. How many children 
were there? | 


29. A is twice as old as B and 6 years younger than O. 
The sum of the ages of A, B, and C is 96 years. What is 
the age of B? 


30. Divide a line 24 inches long into two parts such 
that the one part shall be 6 inches longer than the other. 


31. Two trains travelling, one at 25 and the other at 30 
miles an hour, start at the same time from two places 220 
miles apart, and move toward each other. In how many 
hours will the trains meet? 


te SCHOOL ALGEBRA. 


32. A man bought twelve yards of velvet, and if he had 
bought 1 yard less for the same money, each yard would 
have cost $1 more. What did the velvet cost a yard? 


33. A and B have together $8; A andC,$10; BandC, 
$12. How much has each? 


34. Twelve persons subscribed for a new boat, but two 
being unable to pay, each of the others had to pay $4 more 
than his share. Find the cost of the boat. 


35. A man was hired for 26 days on condition that for 
every day he worked he was to receive $3, and for every day 
he was idle he was to pay $1 for his board. At the end of 
the time he received $58. How many days did he work? 


36. A man walking 4 miles an hour starts 2 hours after 
another person who walks 3 miles an hour. How many 
miles must the first man walk to overtake the second ? 


37. A man swimming in a river which runs 1 mile an 
hour finds that it takes him three times as long to swim a 
mile up the river as it does to swim the same distance down. 
Find his rate of swimming in still water. 


38. At an election there were two candidates, and 2644 
votes were cast. The successful candidate had a majority 
of 140. How many votes were cast for each? 


39. Two persons start from towns 55 miles apart and 
walk toward each other. One walks at the rate of 4 miles 
an hour, but stops 2 hours on the way; the other walks at 
the rate of 8 miles an hour. How many miles will each 
have travelled when they meet? 

40. A had twice as much money as B; but if A gives B 
$10, B will have three times as much as A. How much 
has each? 

41. If 22 —8 stands for 20, for what number will 4—~z 
stand ? 


SIMPLE EQUATIONS. vO 


42. A vessel containing 100 gallons was emptied in 10 
minutes by two pipes running oneat atime. The first pipe 
discharged 14 gallons a minute, and the second 9 gallons a 
minute. How many minutes did each pine run? 


43. A man has 8 hours for an excursion. How far can 
he ride out in a carriage which goes at the rate of 9 miles 
an hour so as to return in time, walking at the rate of 3 
miles an hour? 


44. If 84—4a=22—a, find the number for which 
4x2—Ta stands. 
2% —3a 


r+a 


Norr. When we compare the ages of two persons at a given time, 
and also a number of years after or before the given time, we must 
remember that both persons will be so many years older or younger. 
Thus, if a man is now 2a years old and his son 2 years old, 5 years 
ago the father was 2x—5 and the son x—5, and 5 years hence the 
father will be 2a + 5 and the son a + 5, years old. 


45. If 7x-a=9(«—a), find the number 


46. A man is now twice as old as his son; 15 years ago 
he was three times as old as his son. Find the age of each. 


47. A man was four times as old as his son 7 years ago, 
and will be only twice as old as his son 7 years hence. Find 
the age of each. 


48. A, who is 25 years older than B, is 5 years more 
than twice as old as B. Find the age of each. 


49. A man is 25 years older than his son; 10 years ago 
he was six times.as old as his son. Find the age of each. 


50. The difference in the squares of two consecutive 
numbers is 19. Find the numbers. 


51. The difference in the squares of two successive odd 
numbers is 40. Find the numbers. 


CHAPTER (Vi 
MULTIPLICATION AND DIVISION. 
SPECIAL RULES. 


98. Special Rules of Multiplication. Some results of mul- 
tiplication are of so great utility in shortening algebraic 
work that they should be carefully noticed and remem- 
bered. The following are important: 


99. Square of the Sum of Two Numbers. 

(a+ 6? =(a+ 6)(a+) 
=a(a+b)+b(a+)) 
=a@+ab+ab+0? 
=a+2ab-+ 6’. 

Since a and 6 stand for any two numbers, we have 


RuutE 1. The square of the sum of two numbers is the 
sum of ther squares plus twice their product. 


100. Square of the Difference of Two Numbers. 
(a— 6) = (a— b)(a— 8) 
= a(a— b)— b(a— B) 
=a? —ab—ab+ 0? 


= a — 2ab+4 6. 
Hence we have 


RuuE 2. The square of the difference of two numbers ts 
the sum of their squares minus twice their product. 


SPECIAL RULES OF MULTIPLICATION. 75 


101. Product of the Sum and Difference of Two Numbers. 
(a+ 6)(a— b6)= a(a — 6) + b(a— 6) 
=a—ab+ab—6 
= a — 0’. 
Hence, we have 


Rue 8. Lhe product of the sum and difference of two 


numbers is the difference of ther squares. 


If we put 2a for a and 3 for 6, we have 
Rule 1, (244+ 8? =42?4+ 122+ 9. 
Rule 2, (22— 3f=427 — 122749. 
Rule 3, (2% + 3) (2% — 3) = 42? — 9. 


Exercise 19. 


Write the product of 


L (@+y). 7. (w+y)(e—y). 

2. (%—a)’*. 8. (42 —3)(42+38). 

3. (7+ 2b). 9. (8a?+ 487) (8a — 40’). 
4. (82— 2c)’. 10. (3a—c)(8a—e). 

5. (4y —5)*. 11. (v +78") (a+ 70’). 

6. (8a?+ 42’). 12. (ax+2by) (ax — 2 by). 


102. If we are required to multiply a+d+ce by a+b—e, 
w2 may abridge the ordinary process as follows : 
(a+6+c)(a+6—c)=[(a+6)+e][(a+6)—e| 
By Rule 3, =(a+ b?—¢ 
By Rule 1, =a?+20a0+6—e’. 


76 SCHOOL ALGEBRA. 


If we are required to multiply a+6—c by a—b+e, we 
may put the expressions in the following forms, and per- 
form the operation : 


(a+b—c)(a—b +0) =[a+(b—d)][a—(6—9)] 

By Rule 8, =a —(b—c) 

By Rule 2, = av — (b° — 2c + ¢’) 
=a’— b?+ 2be— ’. 


Exercise 20. 

Find the product of 

1. e+y+z and r—y—z. 
x—y+z and r—y—z. 
ax +by+1 and ax+ by—1. 
1+a—y and 1—z+y¥y. 
a+26— 3c and a—26+3e. 
a—ab+b* and @+ab+ 6". 
mt+mn+n and m’—mn+n’. 
2+e+2? and 2—x£—2'. 
v+ta+l1 and @—a-+l. 
32+ 2y—z and 84—2y-+z. 


CP WH TH P w DV 


a 
~ 


108. Square of any Polynomial. If we put x for a, and 
y +z for 4, in the identity 


(a+ bP =@+ 2ab+ 86’, 
we shall have 
[c+ yt+z) Fae + 2x(y+2)t+ ytz2), 
or (a@+y+2z) =a? 4+ Qay4+ 2az+y?+2y24+ 2 
HeV+Y+e2+ 2ry+ 2a24+ Qyz. 


SPECIAL RULES OF MULTIPLICATION. TT 


It will be seen that the complete product consists of the 
sum of the squares of the terms of the given expression and 
twice the products of each term into all the terms that fol- 
low it. 

Again, if we put a—é for a, and e—d for 6, in the same 
identity, we shall have 


[(a@—6)+(e—a)P 
= (a—b)?+2(a—b)(e—d)+(e—d) 
= (a’—2ab+6?)+2a(ce— d)—26(e—d)+(e —2cd+d?’) - 
= a’?—2ab+67+-2ac—2ad—2be+2bd+ c— 2cd+d’? 
= a? 107+ c+ d?—2ab+2ac—2ad—2be+2bd —2 cd. 


Here the same law holds as before, the sign of each 
double product being + or —, according as the factors com- 
posing it have hke or unlike signs. The same is true for 
any polynomial. Hence we have the following rule: 


Rue 4. The square of a polynomial ts the sum of the 
squares of the several terms and twice the products obtained 
by multiplying each term into all the terms that follow it. 


Exercise 21. 


Write the square of 


1. 2a 3y. 10. Pye’. 

2. at+b-+e. ll. 2xu—y—z. 

3. rty—z. 12. a—2b—38e. 
4. x—y+z. 13. 8a—6+4+2e. 
5. xty+5. 14. «+ 2y— 8z. 
6. 2+2y+3. 15. e+y+2+1. 
7. a—bte. 16. 44+y+2-2. 
8. 82—2y+4. 17. 2e—y—2-—3. 
9. 


24 —3y + 42. (18. c—2y— 8244. 


78 SCHOOL ALGEBRA. 


104. Product of Two Binomials. The product of two bino- 
mials which have the form z+ a, +6, should be carefully 
noticed and remembered. 


(1) (w+5)(@+8)=2(e4+8)+5(e+3) 
=27+821524+15 
= 277+ 844+ 15. 


(2). (2 —b)\(e— 3) =27 (4 3) — 5(¢—3) 
= 2'—38xa2—52415 
= x —82-+ 1. 


(3) (#@+5)(#—38)=2(«—3)+5(#—8) 
= 2?—32+52—15 
=7'?+22—15. 


(4) («—5)(#+3)=2(¢+3)—5 (+8) 
=277+32—5x2—15 
= 27 —2a—165. 


Each of these results has three terms. 
_ The first term of each result is the product of the first 
terms of the binomials. 
The last term of each result is the product of the second 
terms of the binomials. 
The middle term of each result has for a coefficient the 
algebraic sum of the second terms of the binomials. 


The intermediate step given above may be omitted, and 
the products written at once by wnspection. Thus, 


(1) Multiply «+8 by «+7. 
8 ah 10 8 ee D0. 
. (@+ 8) (@+ 7) = 2? + 152 + 56. 


SPECIAL RULES 


(2) Multiply z—8 by x—7. 


(-8)+(-7) =—15, (-8)(-) = +56. 
-. (@— 8)(%@— 7) =2?—15ax +4 56. 


(3) Multiply «— Ty by «+ 6y. 


on QW Or Pp © WD 


eS Se Se Se 
a Ff © Ww =e CO 


6h 


OF MULTIPLICATION. 79 


(—Ty)x6y=— 427’. 
.(a@—-Ty)(@+ by) = 2 — ry — 427’. 
(4) Multiply 2?+ 6(a+ 0) by 2 — 5(a+ 0). 
+6—5=1, 6(a+6)x—5(a+6)=—30(a+6). 
. [24-6 (a+0) | [a5 (a+6)]= a4+-(a+b)x’—80 (a+)’. 


Exercise 22. 


Find by inspection the product of 


- (+8) (4+ 3). 
- («+ 8)(% — 3). 
. («—7) («+ 10). 
- («—9)(%@— 5). 
- (c—10)(x4 9). 
- (a—10)(a—5). 


. («— 8a) (e+ 2a). 
- (a+ 26)(a—40). 


. (a—12)(a—8). 


. (a+ 26) (a4 40). 
- (a—38b)(a+70). 
. (a+ 26)(a— 96). 
. («—38a)(x—4a). 


- («+ 42z)(x— 22). 


- («+ 6y)(e—5y). 


16. 
(a? + 2y?) (a? — 84). 

| (@ $89) (a? —4y/). 

- (ab —8)(ab+5). 

. (ab —Txy)(ab+ 3xy). 

. (w—8y)(e—8y). 

. (+6) (x + 6). 

. (a—8b)(a—38b). 

. (—c)(x—a). 

» («+a)(x—S). 

. («—a)(a#+ 0). 

| [(a-+b)+2]{(a+8)—4]. 
, [(e-+y)— 21 @+y)4+- 4). 
. (e-+y—1) (@+y+10). 
. («—y—T)(#—y—10). 


(x? — 9) (a? + 8). 


80 SCHOOL ALGEBRA. 


105. In hike manner the product of any two binomials 
may be written. 


(1) Multiply 2a—6 by 3a+4+ 40. 
(2a— 6) (8a+46)=6a'+ 8 ab— 8ab —4B6 
= 6a’+ 5ab — 40’. 
(2) Multiply 2x+38y by 3x2 —2y. 
Tke middle term is 
(2x) xX (—2y)+8yX 84=5z2y; 
“. (2e+8y)(82—2y)=62?-+ dry — by’. 


Exercise 23. 
Find the product of 
382—y and 24+ y. 6. 10z—38y and 10% — Ty. 
7. 38a7—20 and 2a’+ 887. 
5a—4y and 84—4y. 8. a+ and a—Od. 
a—Ty and 2e—5y. 9. 38a?— 20? and 2a+4 386. 
lla—2y and 7z+y. 10. a’—B8 anda+6d. 


4x—3y and 3x2—2y. 


o FF © Ww -& 


106. Special Rules of Division. Some results in division 
are so important in abridging algebraic work that they 
should be carefully noticed and remembered. 


107. Difference of Two Squares. 
From §101, (a+ 6)(a— 6)=a@— 0’. 
eo a— oF 


iM =a—b, and =a+b. 
Sab arith ear a+6. Hence 








Rute 1. The difference of the squares of two numbers ts 
dwisible by the sum, and by the difference, of the numbers. 


Write by inspection the quotient of 


1. 


12. 


13. 


SPECIAL RULES 


Cie 
a—2 
Oo aA 
38+ 24 
16) ae 
4ta 
oO. 
xz—bd 
3 dentiat 
6+ 4 
Dor 0" 
eons 




















2527 — 360? 
52+66 


49 ¢7 — ad* 
fod 
Va) 
sat+l 
16—4a@ 
4—2a 


9at — 25y* 


8a°+ 57 
4 oY 
22° — 37° 
4 wl Whey ae 


22° — at 





Exercise 24. 


14. 
15. 
16. 
Rr 
18. 
19. 
20. 
21. 
22. 
| 23. 
24. 


25. 


OF DIVISION. 


aboc® — x 
absc* + x® 
ata — bY 
«at — B° 
Of (Os eye 
a—(b+c) 
a — (86 —4Ac) 
a—(8b—4c) 
1— (Bead a) 
1+ (@—y) 
(Se Bee LOZ) 
(82—y)+ 42 
(BOM) ey 
(7-+3a)—32 
L— (Gee) 
1+ (7a—56) 
(Sa-- Zyl Ae 
(82+ 2y)—2z 
(ae Om (Gay) 
(a— 6) + (¢—y) 
pe (yz) 
a—(y + 2) 
(Seaaye 20 
(c—2y)+5 
(2e-y)i—92t 
(22+ y)—32 


82 SCHOOL ALGEBRA. 


108, Sum and Difference of Two Cubes. By performing 
the division, we find that 


rat +,ab + b%, 








vane = a’?—ab+ 06’, and a’ 


Hence, 


Rute 2. The sum of the cubes of two numbers ts divisible 
by the sum of the numbers, and the quotient rs the sum of 
the squares of the numbers minus their product. 


Rue 38. The difference of the cubes of two numbers is 
divisible by the difference of the numbers, and the quotient 
as the sum of the squares of the numbers plus ther product. 


Exercise 25. 


Write by inspection the quotient of 

















sera a My Oc On 
1— 22 ab —e 
2, } +82 10. ard* + oF 
1+ 22 ab+e 
ai SU aah Oe 11. ae 
3a—b 4+y 
iM 27a + 0 io. 343 — 8a? 
8atb 7—2a 
Oe re ia 21 y* 13)) oe ee 
‘)\ 4a- By eae, 
A 64g + 21y 14: a + 129° 
4a—3y AON ary ask 
1— 272 Oe Aree 
7. ———: 15. 
1l—38z a— 3b 
ena Paley ie Oe Bey 
1+ 8z 2% —4y 


SPECIAL RULES OF DIVISION. 83 
109. Sum and Difference of any Two Like Powers. By 


performing the division, we find that 


CF a8 + 0%b abt $8; 


as 








a Bi sins 3 oth se aot On 








a+6 

z oir — a ad ob*+-0b* + Bt 

(Of 

a+b? 3 3 4 
op ater Wore a ake 2 bs 1 Bt, 

a ‘ 


We find by trial that 
a? + 0?, at+ b*, a& + 0%, and so on, 
. are not divisible by a+ 6 or by a—b. Hence, 
If n is any positive integer, on wb 
(1) a® +0" is divisible by a+b if? n ws odd, and by neither 
a+b nor a—b of n ws even. ee tice 
(2) a*—b” is dwisible by a — bof n ws odd, and by both 
a+b anda—b if nis even. 


Nore. It is important to notice in the above examples that the 
terms of the quotient are all positive when the divisor isa —0, and 
alternately positive and negative when the divisor is a +); also, that 
the quotient is homogeneous, the exponent of a decreasing and of b 

' increasing by 1 for each successive term. 


Exercise 26. 
Find the quotient of 














Phe Arron heirs le | a +32. 
LAY z+1 Peer 
eae iy aaa ae Ok 
x+y x—2 l1—m 
zt—1 a? — 82. : 1+ mm 











; 6. ; 
x—l x—2 l+m 


CHAPTER VII. 
FACTORS. 


110. Rational Expressions. An expression is rateonal when 
none of its terms contain square or other roots. 


111, Factors of Rational and Integral Expressions. By fac- 
tors of a given integral number in Arithmetic we mean 
integral numbers that will divide the given number with- 
out remainder. Likewise by factors of a rational and inte- 
gral expression in Algebra we mean rational and integral 
expressions that will divide the given expression without 
remainder. 


112, Factors of Monomials. The factors of a monomial 
may be found by inspection. Thus, the factors of 147} 
are 7, 2, a, a, and 0. 


118. Factors of Polynomials. The form of a polynomial 
that can be resolved into factors often suggests the process 
of finding the factors. 


Case I. 
114. When all the terms have a common factor. 
(1) Resolve into factors 227 -+ 62y. 


Since 2 is a factor of each term, we have 
Cy ies 


ot Oy oS CY 9 
Q2 Die Qa evanens 
. 2207 + 6a2y = 2x(x+8y). 


Hence, the required factors are 2% and w+ 3y. 


FACTORS. 85 


(2) Resolve into factors 16a*+ 4a?— 8a. 
Since 4a is a factor of each term, we have 


l6a°+4a?—8a_16a* , 4a 8a 


ees es ey 


40, Bota told 
=4¢7+a—2. 
. 16a +4¢0— 8a=—4a(4v+a— 2). 


Hence the required factors are 4a and 4a?+a—2. 


Exercise 27. 


Resolve into two factors : 


1. 38a7—62'. 7. 38a@7b — 4a’. 

2. 2a?—Aa. 8. 82%? + 4277’. 

3. 5ab— 5a’d’. 9. 82t—92’— 62°. 

4. 8a0—a’?+a. 10. 8a?%? — 407) + 12a7y’. 

5. a + ay — ay’. 11. 8ab’c? — 4.a7b°c’. 

6. a’—a’h+a’b’. 12. 15a’x— 10a®*y + 5a*z. 
Case II. 


115, When the terms can be grouped so as to show a common 
factor. 


(1) Resolve into factors ac +ad-+ be + bd. 


ac+ad + be+ bd = (ae + ad) + (be + bd) (1) 
=a(e+d)+b(e+d) (2) 
=(a-+b)(e-+d), (3) 


Nore. Since one factor is seen in (2) to be e+, dividing by e+ d 
we obtain the other factor, a + b. 


86 SCHOOL ALGEBRA. 


(2) Find the factors of ac + ad — be — bd. 
ac + ad — be — bd = (ae + ad) — (be + bd) 
=a(e+d)—b(e+d) 
=(a—b)(e+ a). 
Norr. Here the signs of the last two terms, — bc—bd, being put 
within a parenthesis preceded by the sign —, are changed. 
(3) Resolve into factors 32° —52?— 6x -+ 10. 
32° — 52? — 62+ 10 = (82° — 527) — (6x — 10) 
= 27 (82 — 5) — 2(8a—5) 
= (x? — 2)(8.4—5). 


(4) Resolve into factors 52° — 1l5axv?—x+ 8a. 
5 a§— 15a2°—x2-+ 8a= (52° — 15a2”) — (x — 8a) 
= 52" (« —8a)—1(#—38a) 

= (527 —1)(x— 3a). 


(5) Resolve into factors 6y — 27 2°y —102-+ 452°. 

6 y—27 a’y—1027+45 23 = (452° — 27 2*y) — (102 — 6y) 
= 927 (5x—3y)—2(52 — 3y) 
= (92? — 2) (5x —38y). 

Nott. By grouping the terms thus, (6y — 27 x*y) — (10a — 452°), 

we obtain for the factors, (2 — 9) (3y—5z2). 

But (2 —92?)(3y —5a2) = (9a? — 2)(54—3y), since, by the Law 

of Signs, the signs of two factors, or of any even number of factors, 
may be changed without altering the value of the product. 


: ‘ Exercise 28. 
Resolve into factors : 

5. 2 +ax—bx—ab. 

6. 2+ xy — ax—ay. 
ax — cy — ay + cx. 7. 2—axy—6x+ by. 
2ab—B8ac—2by+3cy. 8 


ax — bz + ay — by. 
ax — bx ~— ay + by. 


m 09 2 me 


. 227—8ay+4ar—6bay. 


FACTORS. 87 


9. wb —abx —ac+ en. 
10. aba + Bex — acy — be’y. 
11. 32° —57? — 62° + 10zy’. 
12. 8az —10b72 —12a+150. 
13. 22°— 382?°— 47-4 6. 
14. 624+ 82° — 922-122. 
15. ax*-+ ba’? — ax — 6. 
16. 8cx*— 2dz> — 9cx’? + bade. 
17. 1+ 152*—52— 82". 
18. awtaaetataz. 
19. (a+b)(ec+d)—3c(a+d), 
20. (c—y)'+ 2y(@—y). 
116. If an expression can be resolved into two equal 
factors, the expression is called a perfect square, and one of 
its equal factors 1s called its square root. 


Thus, 162°? = 4a*°y x 4a*°y. Hence, 162%? is a perfect 
square, and 42°y is its square root. 
Note. The square root of 16 2*y? may be — 4a°y as well as +4a%y, 


for —4a°y x — 4a°y = 16 a*y?; but throughout this chapter the posi- 
tive square root only will be considered. 


117. The rule for extracting the square root of a perfect 
square, when the square is a monomial, is as follows: | 

Extract the square root of the a and divide the 
index of each letter by 2. 


118, In like manner, the rule for extracting the cube 
root of a perfect cube, when the cube is a monomial, is, 

Extract the cube root of the coefficient, and divide the index 
of each letter by 8. 


88 SCHOOL ALGEBRA. 


119. By §§ 99, 100, a trinomial is a perfect square, if its 
first and last terms are perfect squares and positive, and its 
middle term is twice the product of their square roots. 
Thus, 16 a? — 24ab + 98? is a perfect square. 

The rule for extracting the square root of a perfect 
square, when the square is a trinomial, is as follows: 


LEixtract the square roots of the first and last terms, and 
connect these square roots by the sign of the muddle term. 


Thus, if we wish to find the square root of 
16a — 24ab + 90?, 
we take the square roots of 16a’ and 90’, which are 4a 
and 36, respectively, and connect these square roots by the 
sign of the middle term, which is —. The square root is 
therefore ee BF 


In like manner, the square root of 


16a’ + 24ab +96? is 4a+4+ 30. 


Case III. 
120, When a trinomial is a perfect square. 
(1) Resolve into factors z+ 22y + y’. 
From § 119, the factors of 27+ 22y-+ 7’ are 
(e+y(@+y). 


(2) Resolve into factors zt — 22°y + 7’. 
From § 119, the factors of 2 — 2a°y+7? are 
‘ (x? — y) (@—y). 
Exercise 29. 
Resolve into factors : 
1. @&—6ab4+ 90. 3. a—4ab4+ 48, 
2. 4a?+4a6 + 0. 4. 2+ 6ay+97’. 


FACTORS. 89 


5. 427—12ar+ 9a’. 13. 492? — 28 2y + 477. 

6. aw —10ab+ 250’. 14. 1— 206+ 10007. 

7. 4a—4a-+1. 15. 8la’?+126ab + 490°. 

8. 497? — 14 yz + 2’. 16. mn? — 16mna? + 64%. 

9. 2?—1l6z2-+ 64. 17. 4a?— 20az + 252”. 
10. 9a?+ 24ay + 1697. 18. 121a?+ 198ay-+ 81y’. 
11. 16a?+ 8axr+ 2°. 19. a®btc® — Qab?c’a® + x". 
12. 25+ 802+ 642’. 20. 49— 140%? + 100K". 

Case IV. 


121. When a binomial is the difference of two squares. 
(1) Resolve into factors 2? — 7’, 
From § 101, («+ y)(#—y)=2?—y’. 


Hence, the difference of two squares is the product of 
two factors, which may be found as follows : 


Take the square root of the first term and the square root 


of the second term. 
The sum of these roots will form the first factor ; 
The difference of these roots will form the second factor. 


Exercise 30. 


Resolve into factors : 


1. w—4, 6. 25— 16a’. ‘11. 812’?— 47? 
2. 1— 2. 7. 16— 2577’. 12. 64a* — 0. 
3. 2? — 97’. 8. a7b? — 1. 13. mn? — 36. 
4, 4073490") 9x 3° = 100, FA. Lae: 
5. 2? —4y’. 10. 121a’— 360’. 15. 2? — 25. 


90 


16. 49 — 1007’. 
17. 1— 492°. 
18. 4—1217%. 
19. 1— 169a*. 
20. ab? — 4c*. 
21. 9a— a’. 
22. 42% — 7”. 


SCHOOL ALGEBRA. 


23. 
24. 
25. 
26. 
27. 
28. 
29. 


49 ait — y”. 
64a? —9B°. 
81 a‘b* — c*. 
4a°c—9e’. 

20 a®b® — 5ad. 
38a — 12a%e’. 
9a? — 816", 


25 — 647’. 
162" —9ay/. 
25 2” —16 aba’, 
36 aa? — 49 at. 


. 2 —16y’. 
te 4002 
4G ce. 


122. Ifthe squares are compound expressions, the same 
method may be employed. 


(1) Resolve into factors (x+ 8y)? — 16a’. 
The square root of the first term is w+ 3y. 


The square root of the second term is 4a. 
The sum of these roots is e+ 3y + 4a. 


The difference of these roots is 7+ 3y—4a. 


Therefore (« + 3y)?— 16a? =(a#+3y+4a)(a4+3y—4a). 


(2) Resolve into factors a? — (3 6 — 5c)’. 


The square roots of the terms are a and (36 — 5c). 

The sum of these roots is a + (3b — 5c), ora +3b—5e. 

The difference of these roots is a— (3b — 5c), ora—36 + 5e. 
Therefore a? — (36 —5c)? =(a + 3b —5c)(a—3b + 5c), 


1238. If the factors contain lke terms, these terms should 
be collected so as to present the results in.the simplest 
form. r 
(3) Resolve into factors (8a-+ 5b)? — (2a — 3b)". 
The square roots of the terms are 3a +5b and 2a— 30. 


The sum of these roots is (3a + 5b) + (2a— 30), 


or 3a+564+ 2a—3b=54a+4 2b, 
The difference of these roots is (3a + 5b) —(2a— 30), 
or 3a+5b—2a+3b=a+ 8b. 
Therefore (3a + 56)? — (2a —3b)? = (5a + 2b)(a + 8b). 


FACTORS. 91 


Exercise 31. 


Resolve into factors : 


1. @+yy—2. 11. (a—b)’—(e—dy. 

2. (x—y)yr— 2. 12. (2a+ 6) — 25e’. 

3. (~—2y)y?— 42’. 13. (+ 2y)?— (2x2—y/. 

4. (a+ 36)? — 16’. 14. (7+ 3)— (82 — 4). 

5. 2? —(y—2)’. 15. (a+b— c)’—(a—b—ce)y’. 

6. a?— (8b— 2c). 16. (a— 82) — (8a— 22). 

7. B—(2a+ 8c)’. 17. (2Qa—1)?—(8a+1). 

8. 1—(w+56), ~ 18. («—5)—(a#+y—5). 

9. 9a?—(*%— 3c)’ 19. (2a+6—c)—(a—26+c). 
10. 16a@—(2y— 382). 20. (a+2b—8c)?—(a+5c)’. 


124, By properly grouping the terms, compound expres- 
sions may often be written as the difference of two squares, 
and the factors readily found. 

(1) Resolve into factors a? —2ab-+b? — 9c’. 

a —2ab+0?—9e? = (a —2ab+ 0’)—9¢? 
=(a—b)P—9e 
=(a—6b+3c)(a—b—38c). 

(2) Resolve into factors 12ab + 92? — 4a? — 90’. 

Norr. Here 12a) shows that it is the middle term of the expres- 
sion which has in its first and last terms a? and 6?, and the minus 
sign before 4a? and 90? shows that these terms must be put in a 
parenthesis with the minus sign before it, in order that they may be 
made positive. 

The arrangement will be 

92’ —(4a?—12ab +96’) =92?—(2a—3 5)? 

= (32+ 2a—36) (8¢a—2a+ 36), 


92 


SCHOOL ALGEBRA. 


(3) Resolve into factors —a?+8’—c’?+d? + 2ae-+ 2bd. 


Nore. Here 2ac, 26d, and —a?, —c*, indicate the arrangement 


required. 


—a7+h—?+d’?+2ac4+ 2bd 


= (6? + 2bd + d’) — (@ — 2ac + ce’) 
= (+d) ~(a—e) 
=(6+d+a—c)(6+d—a-+e). 


Exercise 82. 


Resolve into factors : 


. @+20b64+0?— 
» &—2ry+y— 9a’. 


- — 2 — oy? — 22y. 
. L—a@&— 2ab— 0’. 


1 5 
2 6 
3. Bb—2#+4ar—A4a’. 7. v@t+b?+ 2ab— 16076". 
4 8 


- 47+ 4064+ 0 — 2’. 
ce 

10. 

iti ts 

12. 

13. 

14. 

15. 

16. 


. 407—9¢0+6a—-1. 
ot Be d?—2ab = 2cd. 

+? — 2ry — 2ab— a — 8’. 
927—62+1—a?—4ab —48. 

a’ + 2ab — 2? — bay — 97’ + OB’. 

a? —9¢4+1—b4 Qby—y’. 
9—62+2?>—a’—8ab—168".. 
4—4¢+2—4a—1—42a’. 

a’ — a —9 +3*+ 6a — 270’. 


125, A trinomial in the form of a*+a’b?+ 6* can be 
written as the difference of two squares. 

Since a trinomial is a perfect square when the middle 
term is twice the product of the square roots of the first 
and last terms, it is obvious that we must add aS? to the 
middle term of a*-+a’b?-+ b* to make it a perfect square. 


FACTORS. 93 


We must also subtract ab’ to keep the value of the 
expression unchanged. We shall then have 


(1) a+ 070? + bt =a‘ + 2073? + bt — 7B? 
fas (a? oe b?)? a2b? 
= (a’+ 0? + ab)(a + b’ — ab) 
= (av + ab + 0) (a? — ab + 8’). 


, if in the above expression we put 1 for J, we shall have 
(2) a@&+t+a@+1=(a+2ae+1)-—a@ 
ages ee (a? ais 1)? ay. 
=(7?+1+a)(?+1-a) 
= (av +a+1)(7?—a-+l1). 


(3) Resolve into factors 42* —372°y’?+ 9y7. 


Twice the product of the square roots of 4a* and 974 is 12a, 
We may separate the term —372y? into two terms, —12a?y? and 
— 25.x7y?, and write the expression 


(4a* — 12.277? + 9y*) — 25277 
= (22° — 387) — 25277 
= (227-38 y+ 5ay) (22°—38y — day) 
= (227+ 52y—3y’) (22°—5ay—3Y). 


Exercise S33. 


Resolve into factors : 


Liat els ye 6. 9a*+ 26076? +.25 O*. 
2. at o?+ 1. 7. 4a*— 21a7y’?+9y*. 
3. 9at—15¢°+1. 8. 4a*— 29 ae + 25%. 
4. 16a°— 17a od} OO Paes SiG atetit- 25 of. 
5. 


4a*—138¢+1. 10. 25at-+ 81a%y?+ 1644. 


94 SCHOOL ALGEBRA. 


CasE V. 
126. When a trinomial has the form x’-+ ax + b.’ 


From § 107 it is seen that a trinomial is often the product 
of two binomials. Conversely, a trinomial may, in certain 
cases, be resolved into two binomial factors. 


127. If a trinomial of the form 2?-++axz+6 is such an 
expression that it can be resolved into two binomial fac- 
tors, it is obvious that the first term of each factor will be 
zx, and that the second terms of the factors will be two 
numbers whose product is 0, the last term of the trinomial, 
and whose algebraic sum is a, the coefficient of x in the 
middle term of the trinomial. 


(1) Resolve into factors «? + 11x + 30. 


We are required to find two numbers whose product is 30 and 
whose sum is 11. 

Two numbers whose product is 30 are 1 and 30, 2 and 15, 3 and 
10, 5 and 6; and the sum of the last two numbers is 11. Hence, 


a2’ + 112+ 30= (x£+ 5) (2+ 6). 
(2) Resolve into factors 2 — 7a + 12. 


We are required to find two numbers whose product is 12 and 
whose .algebraic sum is — 7. 

Since the product is + 12, the two numbers are both positive or both 
negative ; and since their sum is —7, they must both be negative. 

Two negative numbers whose product is 12 are — 12 and —1, —6 
and —2, —4 and —3; and the sum of the last two numbers is —7. 
Hence, 


x’ —Tx+12=(¢—4)(«#—8). 
(3) Resolve into factors a+ 2% — 24. 


We are required to find two numbers whose product is — 24 and 
whose algebraic sum is 2. 


FACTORS. 95 
Since the product is — 24, one of the numbers is positive and the 
other negative; and since their sum is + 2, the larger number is 
positive. 
Two numbers whose product is — 24, and the larger number posi- 
tive, are 24 and —1, 12 and — 2, 8 and —3, 6 and —4; and the sum 
of the last two numbers is + 2. Hence, 


x? + 20 — 24 = (w+ 6)(@ — 4). 


(4) Resolve into factors 2? — 32—18. 


We are required to find two numbers whose product is — 18 and 
whose algebraic sum is — 3, 

Since the product is — 18, one of the numbers is positive and the 
other negative; and since their sum is —3, the larger number is 
negative. 

Two numbers whose product is —18, and the larger number nega- 
tive, are —18 and 1, —9 and 2, —6 and 3; and the sum of the last 
two numbers is —3. Hence, 


2 —3x%—18=(«# — 6) («+ 8). 


(5) Resolve into factors 2’ —10ay+ 97’. 


We are required to find two expressions whose product is 9y? and 
whose algebraic sum is —10y. 

Since the product is + 9y?, and the sum —10y, the last two terms 
must both be negative. 

Two negative expressions whose product is 9 y?, are —9y and — y. 
—3yand —3y; and the sum of the first two expressions is —10y. 


Hence, 
2 — l0xy + 9y' = («—9y)(@—y). 


Exercise 34. 


Resolve into factors: 


1. 2+82+ 15. 4. x —8x—10. 
2. x2?—82+15. 5. + 5ax+ ba’. 
3. a?+2a—-15. 6. w—5av+ 62%. 


96 


SCHOOL ALGEBRA. 


. w —2a—165., 
2+5a+6. 
. &@&—dx2+6. 
. v2@ta—. 

. 2&—a4—6. 

. #+62+ 5. 
. v—6xr+5. 


. v+4ar—5. 
. &—Axr—OSd. 


. 2@+92+ 18. 
. 2 —9x+18. 
. #@+3a—18. 
. #@—38x—18. 
. #&@+9a+8. 
. wv —92+8. 
. &’+Tec—8. 
. @&—Tx—8. 
. &@+724+10. 
. #—Ta+10. 
: x + 3a — 10. 


. «w— dar — 50a’. 


. vy —8ay —4. 
. @—8axr— 5427, 


. af ire — 2c: 


Ga 2B ye + 187 2”. 


32. 
33. 
34. 
35. 
36. 
37. 
38. 
39. 
40. 
41. 
42. 
43. 
44, 
45. 
46. 
47. 
48. 
49. 
D0. 
. vb? — 18 abe 4+ 22’. 
. &@— 1l6ab — 3607. 

- @+172y+ 307. 

. « — ry — 187’. 

. &?+e—20. 

. &+16ab— 26087. 


xv’ + az — 6a’. 


oar —b62*. 


2+ d52y+4y’. 
x? — 3ay —A4y’. 
x — day +4y’. 
xo’ + 3ay—4y’. 
x? —382xy —4y’. 
-a— Tab +1087. 


ax — 8ax — 54, 
wv —Ta— 44. 

xv? + 2 — 182. 

x? — 152+ 50. 

a? — 234+ 120. 
@+17a— 3890. 

c? + 25¢e — 150. 
ce? — 58c-+ 57. 
a*— 11la@’d?+ 800°. 
2’ + 92y + 207’. 
ay? +- 19 xyz + 48 27. 


FACTORS. 99 


7 62-E 724-2. (14. 152? 262x2y + 87’. 
8. 6a? —x— 2. 15. 927+ bry — 8y’. 

9. 152?+142—8. 16. 62?—2y — 35y’. 
10. 82?—10zx+ 3. 17. 102?— 21zy — 10y’. 
11. 1827+ 9a — 2. 18. 142?— 5dxzy + 21y’. 


12. 2427 — l4zy — 57’. 19. 62? — 23 ry + 207’. 
13. 2497 382y+15y* 20. 627+ 352y—6y’*, 


Case VII. 


129. When a binomial is the sum or difference of two cubes, 


at BF 








era S08: Teas OW Cums 
a 
and Cal ab +8; 
= 
“ @ +6 =(a+b)(e?—ab+ 0’) 
and a’ — = (a — b) (av +ab+ 0’). 


In like manner we can resolve into factors any expres- 
sion which can be written as the sum or the difference of 
two cubes. 

(1) Resolve into factors 8a + 276°. 


Since by § 118, 8a*° = (2a)’, and 276° = (807)*, we can 
write 8a’ + 2768 as (2a)* + (36’)*. 


Since | a + 6 = (a+ b)(a — ab + 0°), 
we have, by putting 2a for a and 30? for 6, 
(2a) +(380’)* = (2a+ 30’) (4a°—6ab?+96'). 


100 SCHOOL ALGEBRA. 


(2) Resolve into factors 1252°— 1. 
1252°—1= (62)? —1 
= (52 —1)(252?+ 5%+1). 
(3) Resolve into factors x°+ y’. 
lo = y? = (ao + Weve 
=@ + yah — vy +y’). 
130, The same method is Bee as when the cubes are 
compound expressions. 


(4) Resolve into factors (x — y)* + 2°. 
Since a’+6?=(a+6)(v?—ab+0)"), 
we have, by putting x—y for a and z for 3, 
(w@—yP +2 =[@—-y) +2] [@—-yl—@—y)ete] 
= («@—y+2)(7’—2ay+y—az+y24+2’). 


Exercise 36. 


Resolve into factors : 


1. a’ + 86%. 5. 272*y*®—1. 9. 216a°— 6°. 
2. a&— 27a’. 6. a + 278°. 10. 64.0? = 2706". 
Supa? P64, 7. wy* — 64. 11. 343 — 2’. 
4. 1250+ 1. 8. 64a°+ 1258°. 12. ab? + 348. 
13. 8a*— 0°. 19. 823—(x#—y)*. 
14. 216m’>+ n°. 20. 8(a@+y)+2*. 
15. (a+ 6)—1. 21. 729y* — 642’. 
16. (a—6)+1. 22. (a+ b)>—(a—b)’. 


17. (2a+y)—(x—y)*®. 23. 7290+ 216%. 
18. 1—(a—5)’. 24. xy’ — $122. 


FACTORS. . 101 


181. We will conclude this chapter by calling the stu- 
dent’s attention to the following statements : 


1, When a binomial has the form 2” — y”, but cannot be 
written as the difference of two perfect squares, or of two 
perfect cubes, it is still possible to resolve it into two fac- 
tors, one of whichisa—y. Thus (§ 109), 


a’ — b° = (a— b) (at + ad + 070? + ab? + B'). 


2. When a binomial has the form 2”-+ ¥”, but cannot be 
written as the sum of two perfect cubes, it is still possible 
to resolve it into two factors, except when v is 2, 4, 8, 16, 
or some other power of 2. Thus (§ 109), 


a’ + & = (a+ b) (at — 0°) 4+ ab’ — ab® + DB‘). 


But a+, at+ 5, a®+ 0%, cannot be resolved into 
factors. 


3. The student must: be careful to select the best method 
of resolving an expression into factors. Thus, a®’— 0° can 
be written as the difference of two squares, or as the dif- 
ference of two cubes, or be divided by a — 34, or by a+ 8. 
Of all these methods, the best is to write the expression as 

the difference of two squares, as follows 


(a’)? fed (6°)? —— (a3 +. 6°) (a® meee 6°) 
= (a+b)(a—ab+0’)(a—b)(a*+ab+6’). 


4, From the last example, it will be seen that an expres- 
sion can sometimes be resolved into three or more factors. 


a® — 6 = (a* + b*)(a* — O*) 
= (at +BY (x! +B) (a? — BY) 
= (x* + b*) (a? + 5?) (a + 6) (x — 8). 


102 ’ §CHOOL ALGEBRA. 


5, When a factor occurs in every term of an expression, 
this factor should first be removed. Thus, 
827 —50a?+ 4x —10a= 2(42’ — 25a? + 24 — 5a) 
= 2[ (42? — 25a”) + (22 —5a)]| 
= 2(22—5a)(22+5a+1). 


6. Sometimes an expression can be easily resolved if we 
replace the last term but one by two terms, one of which 
shall have for a coefficient an exact-divisor or a multiple of 
the last term. Thus, 

(1) &—52’?+1]le—15=(2@—52’4+62)+(5x4—16). 

: =2(27—52+6)+5(2—8) 
=(x—3)[x(2—2)+5] 
= (a —3) (#?—24-+5). 


(2) 2—92°+262—24= (2° —92?+14x)+(122—24) 
= 2 (4?—9x+14)+12(4—2) 
=2x(x—7)(«—2)+12(4%—2) 
=(# — 2)(2?—T2+12) 
=(2—2)(2—8) (a—4). 


(3) a — 262 —5=(a — 252) —(4+ 5) 
AG Migs aoe D) 
= (4+ 5) (a — 5x” —11). 


(4) e+32?—4= (2+ 22")+4+ (v’— 4) 
=x’ (x + 2)+ (2? — 4) 
= (x + 2) (2 + # — 2) 
= («+ 2)(4+ 2)(4— 1). 


v 


FACTORS. 


103 


Exercise 37. 


EXAMPLES FOR REVIEW. 


Resolve into factors : 


10. 
| ta 
12. 
13. 
14. 
15. 
16. 
Li 
18. 
19. 
20. 
21. 
22. 


OD WS mT Rw wn 


a’ — 9a. 

xy? — 4 ryt — 3.2°y'. 
vetetat+l. 
v?—Qy+ 2x — zy. 
82° + 2a? — 9a —6. 
xv’ —14x2+ 49. 

36 2” — 497’. 

xt — yt. 
Cay yar be 
z+’. 

xf — , . 
e+ yf. 

x? — (a — b)?. 

mv +2mn+n? — 1. 
a —(m+n)’. 
v—lle+18. 

x’ + 4a — 45. 
e’t+18x2+ 36. 

x’ —138a2— 48. 

x’ + 9x — 36. 
1027+ 2— 21. 

62? —a — 12. 


Lilie ae 


- 122? — x — 20. 

. 9V+12a+4. 

- &—FP—?+ Qobe. 
2 a pal iy’, 

. 2—6z—40. 

. 2 — Tx — 60. 

. @—19a-+ 84. 


. #@+2ar+ 3bx+ Bab. 
Ptr —n?—2mxz. 
. 4a*— 2”. 

. erty”, 

. 9at+ 21 xy? + 25 y4. 
. @—4+y t+ 2xy. 

Me Oe ay 

. 2¢—Ta+ 6. 

. “&—Tev+i. 

. l—-aw— b’?— 2ab. 

. 82°—b62°+4+ 92". 

Silat ee ae eee 2 oid 10). 
. @-tax— ba — ab. 


. 22°—d32ry + 4axr — Bay. 


104 SCHOOL ALGEBRA. 


45. at ba? — aa — b. 63. 6a?—a— TT. 
46. 2+b+a+0. 64. 5ct— 15c® — 90c’. 
47. a&—b+a—Ob. 65. @a—cea+t ary — cy. 


48. (x—y)—2y(x—y). 66. 16z2*—8l. 

49. 1—10zy + 2527’. 67. afta2tl. 

50. a? — B+ Q2be—c’. 68. 272° — 64a’. 

Bl. @t4y7—e—4ay. 69. 2 +47, 

52. a? —407— 9? 412be. 10. 2 —y’. 

53. 42°4+9y—2—l2ay. 71. a’ — 256. 

54. (a+ 6)? — (e— ad)’. 72. «*+ 16072? + 256 a’. 


55. ot y. 73. l1—(@#— yy. 

56. 322°— ce’. 74. (a@+y)+ (22—y). 
57. a&+ 647°. 75. x2® — 216. 

58...729 —a°. 76. 32° +4 — 2. 

BO. tea 77. 2—8a— 2a’. 

60. (a+ 6) — 1. 78. 4—5e— 6c". 

61. @v@—B+a—Qb. 79. 2ry — 2 — y? + 2’. 

62. ata+3b— 98’. 80. 4at'— 9a?+ 6a—1. 


81. a — 2ab+4+ 84 1l2zy—42?— 97. 

82. 22°—42y + 2y’ + 2axr — ay. 

83. (a+ 6)—1l—ab(a+6+1). 

84. att Ba 4 5. (See § 131, 6.) 

85. 62° 2327+ 162—3. (See § 131, 6.) 

86. #+y+2—2ry—222+2yz. (See § 103.) 
87. 4073? — (a+ 0? — e’)’. 


OHAPTER VIII. 


COMMON FACTORS AND MULTIPLES. 


132, Common Factors. A common factor of two or more 
numbers is an integral number which divides each of them 
without a remainder. 


183. A common factor of two or more expressions is 
an integral and rational expression which divides each of 
them without a remainder. Thus, 5a is a common fac- 
tor of 20a and 25a; 327? is a common factor of 122% 
and 152%y. 


134. Two numbers are said to be prime to each other 
when they have no common factor except 1. 


135. Two expressions are said to be prime to each other 
when they have no common factor except 1. 


136. The highest common factor of two or more numbers is 
the greatest number that will divide each of them without 
a remainder. 


137. The highest common factor of two or more expres- 
sions is the expression of highest degree that will divide 
each of them without a remainder. Thus, 3a’ is the highest 
common factor of 3a’, 6a’, and 12a‘; 52’y’ is the highest 
common factor of 102°? and 152°y’. 

For brevity, we use H.C. F. to stand for ‘highest com- 
mon factor.”’ 


106 SCHOOL ALGEBRA. 


To find the highest common factor of two algebraic 
expressions : 


Case I. 


138. When the factors can be found by inspection. 
(1) Find the H.C. F. of 42.00? and 60.a75*. 


A206? 2X 8 XViox aaa x) bbs 
60a b= 2 x2 XO KID x a Dou. 
Spi hev el Uti Bs es Ce OP OUD e 


(2) Find the H.C. F. of 2a@’x+ 2az’ and 8abry + 3 b2’y. 


Qarx + Qaz’ = 2ax(a+2); 
8absy+38bxy = d3bxy(a+2). 
wv the Hy GP) =2(a+ 2). 


(3) Find the H.C. F. of 427+ 42 — 48, 62? — 48a + 90. 
4°+ 4¢4—48 =4(2?+ 2 - 12) 
= 4(x — 3) (x#+ 4); 
62? — 482 — 90 = 6(2? — 82+ 15) 


=6(2 - 3)(e—5); 
Road itso ORLA == 2(#% — 8) 
= 27 -—6, 


Hence, to find the H.C. F. of two expressions: 


Resolve each expression into tts sumplest factors. 

Find the product of all the common factors, taking each 
factor the least number of times it occurs in any of the given 
EXPVeSSloOns. 


COMMON FACTORS AND MULTIPLES. * OF 


Exercise 88. WN 


Find the H.C. F. of 
1., 120 and 168. 4. 36a*2’ and 28 2°y. 
2. 362° and 27 2", 5. 48a7b%c and 60a%e’. 
3. 42a72* and 60 a°z’. 6. 8(a+6) and 6(a+ bY. 
7. l2a(a@+y) and 4b(@-+ y)*. 
8. (vx—1)?(x+ 2) and (x4 — 8) (# + 2)°. 
9. 24070? (a+b) and 42a°b(a-+ 6)’. 
10. x’ (x—-3) and #’—3yx. 12. 2 —4ax and 2’ —62+8. 
11. 2? — 16 and 2’ + 4z. 13. 2? —Tx+12 and 2’ —16. 
14. 92°— 47 and 122? — zy — 6y/’. 
15. #2 —Txe—B8and2?+524+4. 
16. 2+ 32y— 107’ and 2’? — 2zy — 357’. 
17. 2* — 22° — 242’ and 62° —62* — 18027*. 
18. v?— d3ea’y and 2° — 27y’*. 
19. 1+ 642° and 1—42-+ 162”. 
20. «*— 81 and a*+ 82’ — 9. 
21. 2+2e—38, 2 +72+12. 
22. 2?—62+5, 2+ 32 — 40. 
23. 3at+15a*b — 72a’b’, 6a? — 80a) + 86a0’. 
24. bay —122y’°+ by’, 82°y’ + Yay’ — 127%. 
25. 1— 16¢, 14+e— 12. 
26. 827+ 2r—1, 62°+ 724+ 2. 
27. 62? +a—2, 122?—x—-6. 
2h mow Ory Oar) 0a ey Say. 
29. LOnty > ary Cay a oy oye Ay, 


108 SCHOOL ALGEBRA. 


Case II. 
139. When the factors cannot be found by inspection. 


The method to be employed in this case is similar to 
that of the corresponding case in Arithmetic. And as in 
Arithmetic, pairs of continually decreasing numbers are 
obtained, which contain as a factor the H.C. F. required, 
so in Algebra, pairs of expressions of continually decreas- 
ing degrees are obtained, which contain as a factor the 


H.C. F. required. 


140, The method depends upon the following principles: 


(1) Any factor of an expression is a factor also of any 
multiple of that expression. 


Thus, if ¢ is contained 3 times in A, then ¢ is contained 
9 times in 8A, and m times in m A. 


(2) Any common factor of two expressions is a factor of 
ther sum, ther difference, and of the sum or difference 
of any multiples of the expressions. 

Thus, if ¢ is contained 5 times in A, and 38 times in B, 
then ¢ is contained 8 times in A+B, and 2 times in 
A—B. 

Also,in5 A+2 Ait is contained 5x 5+2- 8, or 31 times, 
and in 5 A — 2B it is contained 5 x 5—28, or 19 times. 


(3) The H. C.F. of two expressions is not changed vf one 
of the expressions is divided by a factor that is not a factor 
of the other expression, or uf one 1s multiplied by a factor 
that is not a factor of the other expression. 

Thus, the H.C. F. of 4a7bc? and a’c’d is not changed if 
we remove the factors 4 and 6 from 4a*bc’, and d from 
acd; or if we multiply 4a*bc? by 7, and a’e*d by 11. 


COMMON FACTORS AND MULTIPLES. 109 


141, We will first find the greatest common factor of 
two arithmetical numbers, and then show that the same 
method is used in finding the H.C.F. of two algebraic 


expressions. 


Find the greatest common factor of 18 and 48. 
18) 48 (2 
36 


12)18(1 
12 
6)12(2 
ie 


Since 6 is a factor of itself and of 12, it 1s, by (2), a fac- 
tor of 6+ 12, or 18. 

Since 6 1s a factor of 18, it is, by (1), a factor of 2 x 18, 
or 86; and therefore, by (2), it 1s a factor of 86+-12, or 48. 

Hence, 6 is a common factor of 18:and 48. 

Again, every common factor of 18 and 48 is, by (1), a 
factor of 218, or 36; and, by (2), a factor of 48 — 36, 
prOraLS. 

Every such factor, being now a common factor of 18 and 
12, is, by (2), a factor of 18— 12, or 6. 

Therefore, the greatest common factor of 18 and 48 is 
contained in 6, and cannot be greater than 6. Hence 6, 
which has been shown to be a common factor of 18 and 48, 
is the greatest common factor of 18 and 48. 


142. It will be seen that every remainder in the course of 
the operation contains the greatest common factor sought ; 
and that this is the greatest factor common to that remain- 
der and the preceding divisor. Hence, 


The greatest common factor of any dwisor and the corre- 
sponding duidend is the greatest common factor sought. 


110 SCHOOL ALGEBRA. 


148, Let A and 4 stand for two algebraic expressions, 
arranged according to the descending powers of a common 
letter, the degree of B being not higher than that of A. 

Let A be divided by JS, and let @ stand for the quo- 

ent, and # for the remainder. Then 


BY AQ 
BQ 
R 
Whence, R= A— BQ, and A= BQ+R. 


Any common factor of 6 and # will, by (2), be a factor 
of BQ+ R, that is, of A; and any common factor of A and 
B will, by (2), be a factor of A — BQ, that is, of A. 

Any common factor, therefore, of A and £ is likewise 
a common factor of B and A. That is, the common fac- 
tors of A and B are the same as the common factors of B 
and #; and therefore the H.C.F. of B and J is the 
H.C.F: of A and B. 

If, now, we take the next step in the process, and divide 
B by R, and denote the remainder by S, then the H.C. F. 
of S and A can in a similar way be shown to be the 
same as the H.C.F. of Band R, and therefore the H.C.F. — 


of A and B&; and so on for each successive step. Hence, 


The H.C. F. of any dwisor and the corresponding dwi- 
dend is the H. C.F. sought. 


If at any step there is no remainder, the divisor is a fac- 
tor of the corresponding dividend, and is therefore the 
H.C. F. of itself and the corresponding dividend. Hence, 
. the last divisor is the H.C. F. sought. 


Notre. From the nature of division, the successive remainders are 
expressions of lower and lower degrees. Hence, unless at some step 
the division leaves no remainder, we shall at last have a remainder 
that does not contain the common letter. In this case the given 
expressions have no common factor. 


COMMON FACTORS AND MULTIPLES. 111 


Find the H.C. F. of 227+2—3 and 42°+827’—2-—6. 
227+ 4—3)4a°+ 8a°7— x—6(2r443 


4¢° + 227—62 
62°+ 524—6 
627+ 382—9 
22+ 3)2¢ + 2—3(¢—1 
20-4 32% 
—2x4—3 
Rites ie CO) Milan at 3) ST es 


Each division is continued until the first term of the remainder is 
of lower degree than that of the divisor. 


144, This method is of use only to determine the com- 
pound factor of the H.C.F. Simple factors of the given 
expressions must first be separated from them, and the 
H.C. F. of these must be reserved to be multiplied into the 
compound factor obtained. 


Find the H.C.F. of 
122*-+ 302° — 722” and 322° + 8427 — 1762. 
122* + 802° — 7227 = 627 (227 + 5x4 — 12). 
822° + 842° —176x=42(82? + 21x — 44). 
62? and 4x have 2x common. 
22? + 5x4 —12)82?+ 21¢—44(4 
82? + 202-48 
a+ 4)2e?+5x2—12(22-3 
22° +82 
—38x—12 
.., the H.C. F. = 24 (a + 4). —8xz—12 


bial be SCHOOL ALGEBRA. 


145. Modifications of this method are sometimes needed. 


(1) Find the H.C. F. of 427—8x%—5 and 122°—42—65, 
Aa? — 82%—5)122°— 42—65(3 
199% 24 i hS 


20a — 50 


The first division ends here, for 202 is of lower degree than 42”. 
But if 202— 50 is made the divisor, 42? will not contain 202 an 
untegral number of times. 

The H.C. F. sought is contained in the remainder 20x — 50, and is 
a compound factor. Hence if the semple factor 10 is removed, the 
H.C. F. must still be contained in 2a —5, and therefore the process 
may be continued with 2a —65 for a divisor. 


2%—5)42?7— 84 —5(2r741 


AG? LO ag 
22—5 
22—5 





“thesHvO) Fv 29.5, 


(2) Find the H.C.F. of 
Ola 4a Lor ee and. Ola aD ge 4 ne 
212° — 42° — 15x — 2) 21 2° — 322°— 54x—7(1 
Wa Ae Sy ee 


— 2827 — 39x%—5 


The difficulty here cannot be obviated by remowng a simple factor - 
from the remainder, for — 28a?— 39a%—5 has no simple factor. In 
this case, the expression 21 a3 — 4a?— 15% — 2 must be multiplied by 
the simple factor 4 to make its first term exactly divisible by — 28 a?. 

The introduction of such a factor can in no way affect the H.C.F. 
sought, for 4 is not a factor of the remainder. 

The signs of all the terms of the remainder may be changed; for 
if an expression A is divisible by — F, it is divisible by + F&. 

The process then is continued by changing the signs of the re- 
mainder and multiplying the divisor by 4. 


COMMON FACTORS AND MULTIPLES. Ets 


282? + 3894+ 5)842*?— 162°?— 60x2— 8(82 
842° +1172?+ 152 
—1832?— T5x— 8 
Multiply by —4, —4 
532 2° + 38002 + 32(19 
538227 + 7412+ 95 
Divide by —68, 7703) aa lie 00 
(eee at 
Tx + 1)28a?4 392+4+5(42+5 
2827+ 42a 
s0z2+5 
Parnes [Cs a==11 ae, 1 B0z¢2+5) 


(3) Find the H.C. F. of 
82? + 24 —838 and 62° + 52? — 2. 
627+ 5a?— 2 
4 
82? + 24 —8)242°4 20z7— 8 (84+7 
2427+ 62?— Ix 
1427+ 9x— 8 
Multiply by 4, 4 
562° + 3862 — 82 
562° + 142 — 21 


Divide by 11,  —«:11) 22@—11 
‘ 2%— 1)8274+22—3(42+3 


S2°—4 a 


Gro 
Votre: Ha 2 ae 1, 6z2—3 





114 SCHOOL ALGEBRA. 


The following arrangement of the work will be found 
most convenient: 







827+ 224-3 |-628-+ 527— 12 
82? — 4a 4 
62—3 2427+ 2027— 8 OX 
62 —3 242° + 62°7°— Ie 
1427+ 9x2-— 8 
4 
5627+ 36a — 82 +7 


5627+ 142 — 21 


11)22¢—11 
2%— 1 4z4+38 


146. From the foregoing examples it will be seen that, in 
the algebraic process of finding the H.C.F., the follow- 
ing steps, in the order here given, must be carefully 
observed : . 

I. Simple factors of the given expressions are to be re- 
moved from them, and the H.C. F. of these is to be reserved 
as a factor of the H.C. F. sought. 

II. The resulting compound expressions are to be ar- 
ranged according to the descending powers of a common 
letter; and that expression which is of the lower degree is 
to be taken for the divisor; or, if both are of the same 
degree, that whose first term has the smaller coefficient. 

III. Each division is to be continued until the remainder 
is of lower degree than the divisor. 

IV. If the final remainder of any division is found to 
contain a factor that is not a common factor of the given 
expressions, this factor 1s to be removed ;. and the resulting 
expression is to be used as the next divisor. 

V. A dividend whose first term is not exactly divisible 
by the first term of the divisor, is to be multiphed by such 
a number as will make it thus divisible. 


OO i a ce At eames me 


wD wD WM WD S&S BSF BSB SB BP SYP BS SBS SS 
Cian ser i 80s OO oO OR ON OS 


COMMON FACTORS AND MULTIPLES. 115 


Exercise 39. 


Find by division the H.C. F. of 


CHL ale ete me NON Seer ay Uae sins ea BB 

20? —6a'+ 5a—2, 829 23a°+172—6. 
2027 + 22? —1824+ 48, 202*— 1727+ 482 —8. 
4a3 —227—16x2—91, 122° — 282? — 3872 — 42. 
120 +4e¢°+172—838, 242° —52e°?+ 1l4a—1. 
2¢ + 52? —-244+8, 82°44 2e?—17x+ 12. 
82*— 62° — 27? 1+152¢—25, 4° +72? — 382 — 15. 
4e?—4a°—52+8, 102?—192-+ 6. 

6c —182°+ 382°+ 27, 62*—102°+ 42? — 62+ 4. 
20°82 +2 —2e—38, 4a° tar t4n— 3: 
3824 — xv —2e°4+ 22-8, 62°+182?+ 324+ 20. 


» 82°4+2e'+ 27, 82°+22? —3874+ 27-1. 

. 62°—92*+1]l2*+6 2?—102, 42°+102'+10 274-4 2*160 2. 
. 22°—11l2’—9, 42°4+112*+ 81. 

. 2—4e+ 102? —1927+9, 2+ 22°+ 9. 

. 22°—82?—1627+24, 4a°+ 22* — 28 2°? — 162’ — 822. 
. @—e—l4e’?+e4+1, ve —42*-—# —2e4+ 824 2. 
. 62?—14a2?+ 6e7r—4 a8, x*—ax?— ae? — aa — 2a. 

. 2a'—2e0°—38a— 2a, 8a'—a’?—2a?— 16a. 

. 2a + Tar? + 4a’%x— 3a’, 42° + 9az’? — 2x — a’. 

. 22° —9av’?+ 9a'a — Ta’, 42° — 20az?+ 20¢2—16 a’. 
. 2¢44+-92? + 1427438, 824+ 142749242. 

~ 92° — 72? + 8a? + 2a—4, 62*—Ta?—102?4+5242. 


116 SCHOOL ALGEBRA. 


147. The H.C. F. of three expressions may be obtained 
by resolving them into their prime factors; or by finding 
the H.C. F. of two of them, and then of that and the third 
expression. 

For, if A, B, and C’are three expressions, 

and D the highest common factor of A and B, 

and 7 the highest common factor of D and C; 
Then D contains every factor common to A and B, 

and # contains every factor common to D and C. 
.. # contains every factor common to A, B, and C. 


Exercise 40. 
Find the H.C. F. of 
e’t+38e24+2, 2+424+3, 2?+62+5. 
x’ —9x2—10, 2? —Tx—30, 2—llx+10. 
v—1, #—227?+1, #-—22+1. 
62°+a—2, 2a°+Tx—4, 2e?—Te+ 8. 
vt+2ab4+ 0, &— 8, a +207) + 2ab? + 6°. 
xv —5art+4a?, 2 —8ar+2e, 3x27°—10axr+ Ta’. 
vta—sb, 2—2e—2+2, 2’?+82?— 62-8. 
e+Ta’+ 5ba—l, 24+ 382—32°—1, 382°4+ 5e’?+e—1. 
2—62?+1le—6, 2°—8274+192—-12, 2®§—927?+ 264—24. 


148. Common Multiples. A common multiple of two or 
more numbers is a number which is exactly divisible by 
each of the numbers. 

A common multiple of two or more expressions is an 
expression which is exactly divisible by each of the ex- 
pressions. Thus, 48 is a common multiple of 4, 6, and 8; 


48 (x? — y’) is a common multiple of 3(a—y) and 8(x+ y). 


COMMON FACTORS AND MULTIPLES. LF 


149, The lowest common multiple of two or more numbers 
is the least number that is exactly divisible by each of the 
given numbers. 

The lowest common multiple of two or more expressions ~ 
is the expression of lowest degree that is exactly divisible 
by each of the given expressions. Thus, 24(2?—7’) is the 
lowest common multiple of 8(a@—y) and 8(#+y). 

We use L.C.M. to stand for “lowest common multiple.” 

To find the L.C. M. of two or more algebraic expressions : 


(lar hy 


150, When the factors of the expressions can be found by 
inspection. 
(1) Find the L.C.M. of 42.070? and 60a70*. 
BOO ORAZ CO RLS nn 
60a =2X2X3xK5xXa x Ut. 


The L.C.M. must evidently contain each factor the greatest num- 
ber of times that it occurs in either expression. 


aly Or Ms oe 2s ee Xa OS Be, 
a= 4 G05, 
(2) Find the L.C.M. of 
4a? + 4 —Xbey 6a°—48a2+90, 42?—10%4—6. 
4¢+ 47—12—4(4'4+2—TR) =2x ate —3)(e#+4); 
62?—48 2+ 90=6(2?—82+15)=2 x 3(4— 8) (x— 5); 
4?—lO0e— 6=2(22'—52—3) =2(e#—3)(22+1). 
6 L.0.M.=2x 2x 8 xX (#@— 3) (#+4)(«—5)(22+1). 
Hence, to find the L..C. M. of two or more expressions : 


Resolve each expression into its simplest factors. 

Find the product of all the different factors, taking each 
factor the greatest number of tumes rt occurs in any of the 
gwen expressions. 


118 SCHOOL ALGEBRA. 


Exercise 41. 


Find the L.C.M. of 
1. 24, 82, and 60. 
24072’, 60a%2", and 32a72’. 
x’? —2aey+y* and 2’ —y7’. 
v—4xr4+4, 2?+4r74+4, and 2’? —4., 
s+a and 2?—a?’. 
V’vtart+a’, 2—a’, and z*—a’. 
x? (%#— 3) and 2?— 52+ 6. 
v’+Tx+12 and 2*— 92’. 
se—T2+10, 2—42¢—5, 2?—2«— 2, 
1—382—42?, 1—42—52’, 1—92— 102”. 
62°+ Tay -3y’, 82°+ llay—4y’, 22°+ 1llazy+12y7’. 


er =) 
wo 


12. 8—14a+6a’, 4a+4a’?—8a’, 4¢7+2a0°—6a%. 
13. 62°+72?— 32a, 382°+142—5, 62°+38927+45. 
14. 6ax+9bx—2ay—3Sby, 62°+ 38axr— 2xy—ay. 
15. l2Zax—9ay—82y+ by’, bax+ 38ay—4ay—2y’. 
16. 272°—a', 62°+ax—a2, 152°?9—5ax+3br—ab. 
17. 2 —1, 22? -—a—1, 82’?—a4—2. 


CASE Ii; 


151, When the factors of the expressions cannot be found by 
inspection. 

In this case the factors of the given expressions may be 
found by finding their H.C.F. and dividing each expres- 
sion by this H.C. F. 


COMMON FACTORS AND MULTIPLES. 119 


Find the L.C. M. of 
62 — 1l1lzy+2y and 92 — 2277-87. 


62-llry 427 | 92-227 87 3 
62— 8ry4z2/7 2 


— 3ryttry-iy 13 ope apd ) 
— 8ryttafi2y |182-332¥t 67 
lly)332y—4t2ry_-2y 
oz — 4zy— 27 |2r—y 
.. the H.C_P.=—327 —4zy— 27’. 
Hence, 62° —l1xy+2y=(22—y)(32—42y—2y), 
and 92 — 222 — 877 = (82-+4y) (3 —42y—27/). 
-. the L.C.M. =(2z—y) (82-4 4y) (82°—42y—2y/). 


Exercise 42. 
Find the L.C.M. of 
1. 62°— Tar — 20a@zr, 32° + ar—4a. 
2. 32°—13r + 32-21, 6f4+7r-4H42+21. 
3. 32 —32y7+ 2/7 —y, 42—xy— 3x7. 
4. o—2eéte 24¢—2&—2e¢-—2 
5. +—82r+3, *#—3r+ 2lzr—8. 
6. @—6a*r+ l2ar—S8r, 2a°— Sart 8z. 
7. 22° +2 —122+4+-9, 22 — 727 -+122—39. 
8. Tx —22e—5, TF + 127+ 10r+5. 
9. 2—138r+36, f—xYr—Trtets. 
10. 22°+32x°—Txr—10, 4 —4r—9r+5. 
ll. 122°—2x°—302— 16, 6 —2e— 132—6. 
12. 62 +2—52r—2, 62 +5r—32r—-2 
13. ©—9r+ Mr, SP-—Lrttiz— oo. 


CHAPTER IX. 


FRACTIONS. 


152, An algebraic fraction is the indicated quotient of 


5 : b a 
two expressions, written in the form a 


The dividend a is called the numerator, and the divisor 0 
is called the denominator, 

The numerator and denominator are called the terms of 
the fraction. 


153, The introduction of the same factor into the divi- 
dend and divisor does not alter the value of the quotient, 
and the rejection of the same factor from the dividend and 
divisor does not alter the value of the quotient. Thus 
Bees See 3, alt) NE follows, therefore, that 
4 2x4 4-2 : 

The value of a fraction is not altered of the numerator and 
denominator are both multyphed, or both divided, by the 
same factor. | 








REDUCTION OF FRACTIONS. 


164, To reduce a fraction is to change its form without 
altering its value. 


Case I. 


155. To reduce a fraction to its lowest terms. 

A fraction is in its lowest terms when the numerator and 
denominator have no common factor. We have, therefore, 
the following rule : 


FRACTIONS. LOR 


Resolve the numerator and denonunator into ther prume 
factors, and cancel all the common factors; or, diwide the 
numerator and denominator by ther highest common factor. 


Reduce the following fractions to their lowest terms : 


(1) eCao en 2 x ido oe 1 abi 





5sTabe? 8x 19abe? ~—- Ba 

Cee eae) Aa eta 7) GE ee 
a — 2 (a—a)(a+2) ata 

(3) CH Ta+10_ (@+5)(@+2) +5 
@+5a+6 (a+3)(a+2) a+8 

(4) OG saa Ore (40a 5) (80 To) Soe 
Ba! —2a—15  (2e—8)(4e+5) 4a+5 

(5) cee eae ion 

v—22°+4x2—3 


We find by the method of division the H.C.F. of the 
numerator and denominator to be x — 1. 

The numerator divided by «—1 gives 2? — 32+1. 

The denominator divided by x—-1 gives x —a +38. 


pe — ee Be a Be 


Og ESR 5 ae Re Ae SN eee eats 











Exercise 43. 


Reduce to lowest terms : 

















1 6 ab? A 42m'b- 7 34 any” 
" Garb " 49 mn? " Blatay’ 
oan abrc 5. 30 it ae 9. oo abic? 

15a°b’c? 182°472 5 a®bie 
3 28 ee g, zimin’ g, D8abict 











39 xy " 28 mp " 87 abe 


i) 


12. 


13. 


14. 


15. 


16. 


17. 


18. 


SCHOOL ALGEBRA. 


Jay — lay? | 
122° — l6zy 
eae a 
4a’ + 6ae 
3a*+6a __ 
v+4a+4 
b— 56 
6°?’— 46—5 

20 (a — ¢*) 
4(a?+ ae + c’) 
Dial aa ae 
“+ 22y+y' 

eA 
e+ 2x2—15 
x? —82+15 
227— 132+21 
er ee 
227—Tx—15 


19. 


20. 


21. 


22. 


23. 


24. 


25. 


26. 


27. 


4x°+ l2az+9a° 
82° + 27a? 


v—y - 2yz— 2" 

V+ 2ry+y—-—? 

e+ a’y*? + yf* 
oy : 

2a7 + lWa+21 

3a@’+ 26a-+ 35 


(a+ 6)? — 
(a+6+ cy 
anor Yi 
Y—x 
(oa)? Soe 
(c+ bP—a@ 


(GERD) ea) 
(a+c)— (64d)? 


(atecyr—b 
Age? — (a? + ae: by 


Reduce by finding the H.C.F. of the terms: 


28. 


29. 


30. 


31. 


i ee 
82°—82+8 


Die to 








e+ 42°—5 


2a meats oe a 
82° —42°7—ax2+2 


z'— 1327+ 86 


a — 2 —Taettat6 


32. 


33. 


34. 


35. 


82°+1ia'+22¢7+8 
62° +25 2+ 2382+ 6 


Pe lad aes 
e+ T+ 5a—25 
PN Riera ll eee ete 
82+82r7+a—2 


a 4 DEAS Ae 
w—xe+8ec-—8 


FRACTIONS. 123 


Case II. 


156. To reduce a fraction to an integral or mixed expression. 





to an integral expression. 


(1) Reduce a 
a 


3 


—l_ - 
—~=a+a+1. (§ 108) 








b] 
(2) Reduce rai to a mixed expression. 











Gott ee et We ae 
vte #—-at+l1 
—2’—] 
—27—x 
x—l 
a+] 
—2 
zve—l 2 
ed jess 
w+ rane x+1 


Nors. By the Law of Signs for division, 
—2 2 2 


and = é 
e+] —(x% +1) e+] 











The last form is the form usually written. 


157. If the degree of the numerator of a fraction equals 
or exceeds that of the denominator, the fraction may be 
changed to a mixed or integral expression by the following 
rule: 


Divide the numerator by the denominator. 


Nore. If there is a remainder, this remainder must be written as 

“the numerator of a fraction of which the divisor is the denominator, 
and this fraction with its proper sign must be annexed to the integra] 
part of the quotient. 


124 SCHOOL ALGEBRA. 


Exercise 44. 


Reduce to integral or mixed expressions: 





1 Ae ye 8 2G ey eee. 
4x > a+b 
2 See tee ce 9 Stesieee ark 
32 x+4 
Sa aka Aes TOs seer 
Le ay a+2ux 
pcan an Ne eo Oe, 
ey y 2-4 
pees Te ee 
Ly x’ +x4—-3 
eon 13, 42 t6ar+9a? 
“ey 20 oo 
eee 14 eee oe 
z+ 3 a’+a—2 
Casx III. 


158. To reduce a mixed expression to a fraction. 
The process is precisely the same as in Arithmetic. Hence, 


Multiply the integral expression by the denominator, to 
the product add the numerator, and under the result write 
the denominator. 7 








(1) Reduce to a fraction ~— “+ Les Dd. 
OS ee img a ace) tee ee) 
x—A4 gmwe 4 
_2—8+2°—94+20 
2 4. 
ee rd ae 


xa—A4 


FRACTIONS. 125 


ce re ODO 
a-tb 
a—b —2 = =H _(a— 4) (a+ 6) — (a? — ab — 0") 
a+b a-tb 
a Oe et ad 
a+b 


(2) Reduce to a fraction a—6b 


ab ; 
até 


Nors. The dividing line between the terms of a fraction has the 
force of a vinculum affecting the numerator. If, therefore, a minus 
sign precedes the dividing line, as in Example (2), and this line is 
removed, the numerator of the given fraction must be enclosed in a 
parenthesis preceded by the minus sign, or the sign of every term of 
the numerator must be changed. 





Exercise 45. 


Reduce to a fraction : 


























Se x—3 
ae 
ofl Se. 9. a nt 
3 
De ee 10. &@+art 2’ — aad 
2 a—382 
ee 11, ———— — 20 
jie a5 4 a+2x 
ey ac? Is slag eared heme 
a+b a+b 
5a Loa 13. 22—7 dhe 130 
a x—3 
a+ x ies 14. er age tied 


126 SCHOOL ALGEBRA. 


CasE IV. 


159. To reduce fractions to their lowest common denominator. 


Since the value of a fraction is not altered by multiply- 
ing its numerator and denominator by the same factor 
(§ 153), any number of fractions can be reduced to equiva- 
lent fractions having the same denominator. 

The process is the same as in Arithmetic. Hence we 
have the following rule: 


Find the lowest common multiple of the denominators ; 
this will be the required denominator. Duwide this denomi- 
nator by the denominator of each fraction. 

Multiply the first numerator by the first ieee the sec- 
ond numerator by the second quotient, and so on. 

The products will be the respective numerators of the 
equwvalent fractions. 

Notr. Every fraction should be in its lowest terms before the 
common denominator is found. 


aS 
(1) Reduce io a and oo to equivalent fractions 


having the lowest common denominator. 


The L.C.M. of 4a?, 3a, and 6a’ = 1243. 
The respective quotients are 3a, 4a?, and 2. 
The products are 9a, 8a?y, and 10. 

Hence, the required fractions are 


9ax 8 ary ea ba Le 
12a? 12a? 12a 








(2) Reduce 65 ALA We 30 Or aoe to 
V’+5e4+6 v4+474+8 #4+2r4+1 
equivalent fractions having the lowest common denom- 
inator, 


FRACTIONS. 
net LOU Mrs cai aga nN CSS Ue) 
2+574+6 27449743 a? +2241 
1 ee 3 


— @+3)(@4+2) @+3)\@4+1) @+1I)@FH 
“. the lowest common denominator (L.C. D.) is 
(w + 3)(@ + 2)(@ + 1) (a + 1). 
The respective quotients are 
(«© +1)(e@ +1), (w@ + 2)(~ +1), and (# + 3) (x + 2). 
The respective products are 


L(x +1)(@+1), 2(@+2)(x+1), and 3(a43)(a + 2). 


Hence the required fractions are 


(x + 1) (a + 1) 2(x% + 2)(@ + 1) 
(x + 3)(@+2)(e@ +1)? (@+3)(@ + 2)(x +1) 


3 (x + 3) (a + 2) 
(x + 3)(x% + 2)(x + 1)? 
Exercise 46. 


Express with lowest common denominator : 






































127 


a—22 32° — 2ax 4 4¢7+¢ 2a-e 
Sa Jax Nd greet V9 oie 
tbe ot NN ae alt Sted ee 

e++2 £43 2547—4y? d4+2y 

a a 6 les wr eg 
t-~a 2—@a@ PEO eo 
7 ie LE eelath, 2 
se aah eee p eae a 
8 1 1 3 
" 142e% 1-427 1-22 
9 5 : 7 , 3 
ee EE Bye 
10. L E 


v—9x+18 2?—10x2+ 24 


128 SCHOOL ALGEBRA. 


ADDITION AND SUBTRACTION OF FRACTIONS. 


160. The algebraic sum of two or more fractions which 
have the same denominator, is a fraction whose numerator 
is the algebraic sum of the numerators of the given frac- 
tions, and whose denominator is the common denominator 
of the given fractions. This follows from the distributive 
law of division. 

If the fractions to be added have not the same denomi- 
nator, they must first be reduced to equivalent fractions 
having the same denominator. (§ 159.) 

Hence, to add fractions, we have the following rule: 


Reduce the fractions to. equivalent fractions having the 
same denominator; and write the sum of the numerators 
of these fractions over the common denominator. 


161. When the denominators are simple expressions. 


‘ -- 8a@—46 2a—b+te,a—4e 
1) Simplify ———— — ———— + ——.. 


The L.C. D. = 12. 

The multipliers, that is, the quotients obtained by dividing 12 by 
4,3, and 12, are 3, 4, and 1. 

Hence the sum of the fractions equals 


9n42120) Sa 42o44e%0—46 





12 12 be 
_ 9a—126—(8a—4b+4c)+a—4e 

es 12 
_9a—12b—84a+4b—4ce+a—4e 

12 
2a — 8b = 56 
12 
a—APa Ae 


FRACTIONS. 129 


The above work may be arranged as follows: 


The L. C.D. = 12. 
The multipliers are 3, 4, and 1, respectively. 


3(3a—46) = 9a—126 = lst numerator. 

—4(2a—b+c)=—8a+ 4b—4c¢= 2d numerator. 

1(a — 4c) = ls '@ —4¢= 3d numerator. 
2a— 8b—8e 


or 2(a—4b—4c) =the sum of the numerators. 


: . 2(a—4b~-4c) a—4b—4e 
. sum of fractions = eee eros 
12 6 


Exercise 47. 


Simplify : 


Tc aes mar AI as A acca 














Tx<—5 8x2+2,2+1 52z—-10 
CN Sa) eg can RG ei Rn lc 
8 o T 4 i 


22+3,%2-2 54+4 2xr-—4 
9 8a) 12 3 


2e73, 248 182#+5 «-88 


22 4% 82 x 








2 3 ars wy on 


2 2 2 RIL Ae 
4a She WA ets Ae 4at 








6? ab a ab? 


Dice Agee ln wld ee Ne eB 


4 10 ee 3 


Crk Aan Sk OD RApi, 
2 ee 9) A 8 











180 SCHOOL ALGEBRA. 














2x—6 8x—4 , 56x—48 
10. — —__——- 
5x 15% 5 45 x 
‘ie . a igs Oe ee 
oy xy” x'y 
3 it 1 2E— 6 Yo OZ 
Wag pe, Ie Be 
EI 6 yz eile i ar A xyz 


162. When the denominators have compound expressions. 





—b até gt 
The L.C. D. is (a — b)(a + 3). | 
The multipliers are a + b, a— 6, and 1, xespectively. 
(a+ b)(2a+b)= 2a*?+3ab + b? = Ist numerator. 
—(a—b)(2a—}b) =— 2a? + 3ab— b? = 2d numerator. 
—1(6ad) oe —6ab = 3d numerator. 


0 = sum of numerators. 





» Q 


.. sum of fractions = 0. 











: tA SO 
2) Simplify = . 
G) Pay Posy pig Ona 
The L.C. D. is (a — 2)(« — 3) (a — 4). 
(x —1)(e~—3)(e@—4)= a ~— 8a? +192 —12 = Ist numerator. 
(w — 2)(v— 2)(a@—4)= 2 — 82? + 20%—16 = 2d numerator. 
(a — 2)(a—3)(@—3)= a— 8a? + 212 —18=3d numerator. 
323 — 242? + 602 — 46 = sum of numerators 
* sum of fractions = 30° — 242? + 60% — 46. 
(« — 2)(« — 3)(e—4) 


Exercise 48. 








Simplify : 
1. ar ter 3. 8 bea) 
xrt+6. «—5 ltz 1-2 
ag Siatentesieny eL2a) P icba sok hat Near 


1 ak ee Dir ape ayy 


FRACTIONS. 


Th iccsen hand Caan’ BE 
Be, Ve tuys 
oe aaa 
a(a+b) 2a(a—6) 
St Nia ek ad read 
elt ate 
AND C0 88 0) 
a= 86 BITS 
1 


7 13. 
aieSye 





14. 


15. 





16. 


i (A 
of 





Eo 


oy aN al eB 
Doar ee ARI), 





131 
ary x~y_ 4ay 
Ct) ee 
AGN Oi, 
ata a@+2 


1 


10. 








11. 











12. 


—— eee 


1 








(a+ 6)° 
we ee Ye 


Hint. Reduce the last fraction to lowest terms. 


18. 3 





1 
x— 2a 


19. 





20. 
z+] 
a1, ator. 


22. 


4a 


L— oO (c—a)? (e—a) 


v’—2x2+38 


ape) Peetu kit Y 
2-38 #w+3824+9 
@+8xe2+15 «ex—1 
V’+tTx<+10 «¢—2 


5a? 


aa xta 


—8a w+2axrt4a* 


x 


eimai a 
v’—xtl «£41 





o+a—3 
ur To aT 





Hint. Reduce the first fraction to lowest terms. 


Vea SCHOOL ALGEBRA. 


Fi SO OG Nae 
2’—8ar+15¢7 x«—5a 
1 Nene 
e—-2 w#—82e+2 w#—427+8 
Hint. Express the denominators of the last two fractions in 
prime factors. 


23. 





24. 


25. at AE SAAR PU se DS BE Bk 
@—-Ta+tl1l2 @¢v—4a+8 ad—5a+4 

Py peat cal te a) 
l0?+t+a—8 2¢+7T7a—4 

27. 3 | 


yn aT ys Ree 


168, Since ab _ a, and ab a, it follows that 


b —b 
The value of a fraction is not altered uf the signs of the 
numerator and denominator are both changed. 
It follows, also, by the Law of Signs, that 
The value of a fraction is not altered if the signs of any 
even number of factors in the numerator and denominator 
of a fraction are changed. 


164. Since changing the sign before a fraction is equiva- 
lent to changing the sign before the numerator or the 
denominator, it follows that 


The sign before the denominator may be changed, provided 
the sign before the fraction 1s changed. 


Note. Ii the denominator is a compound expression, the beginner 
must remember that the sign of the denominator is changed. by 
changing the sign of every term of the denominator Thus, 


wv x 


-a— 2 ~—a 





These principles enable us to change the signs of frac- 
tions, if necessary, so that their denominators shall be 
arranged in the same order. 


FRACTIONS. 133 


3 22—8 
eee 
22—1 1—42 
Change the signs before the terms of the denominator of the third 
traction, and the sign before the fraction we have 





Gy sate a 
xv 


The L.C. D. = #(2%—1)(2a + 1). 
2(2e@—1)(24+1)= 8a?—2 = 1st numerator. 
—3a(22+1) =— 62?— 3a = 2d numerator. 
—a(22—3) =—227+3a2=3d numerator. 
—2 =sum of numerators. 
. sum of the fractions = — ili Si leo 
x (2%—1)(2x% +1) 
(2) Simplity 
1 1 1 
Te Se SE ae ee -- ee ee ae + | SS EEE SE ES 
a(a—b)(a—e) 6(6—a)(b—€¢)~-c(c—a)(e—)) 
Norr. Change the sign of the factor (6 —a) in the denominator 
of the second fraction, and change the sign before the fraction. 
Change the signs of the two factors (e—a) and (c— 8) in the de- 
nominator of the third fraction. We now have 
LS GLUT hes eg SAN Ese 
a(a—b)(a—c) b(a—b)(b—c)~ c(a—c)(6—¢) 
The L.C. D. = abe(a — b)(a — c)(b —¢). 


be (b — c) = b?c — be? = Ist numerator. 
—ac(a—c)=—ae+ae = 2d numerator. 
ab(a—b) = ab — ab? = 3d numerator. 


a®b — a?c — ab? + wc? + b%ce — bc? = sum of numerators. 
= a?(b —c) — a (0? — c?) + bc (b— 0), 
=(a?—a(b +c) + be][b— cl], 
=[a? — ab — ae + be)}[b — ¢], 
= [(a? — ac) — (ab — dc)][d — ¢}, 
=[a(a —c)— d(a—c)][b —c], 
= (a — b)(a—c)(d— Cc). 
- sum of the fractions = _(a—b) (a — 0) (be) 1 


abe(a — b)(a — c)(b —c) ~ abe 


184 


SCHOOL ALGEBRA. 


Exercise 49. 








Simplify : 
x? x x 
i Het is 
Ser rarant 1—z 
9 a 3a 2ax é 
az ata2 x#-—-@ 
a 3 ye 15 


10. 


i Mite, 


12. 


13. 








On) 8 WS 499 GeO 4 oP 


a—b 2a a’ -+-a’b 
b PAG, ea igen 


3 5 ed t 





























eh eo 7 Manat TY 
1 1 1 
(ecient a eae 2 — at 
1 spe ae NA ee gee 
e—y @+oayty yx 
] oy i| 
CT CR CIEL Ey ETO ERC 
be ac ab 
(oO GED) Dees 
b+te ate Ls ee! es Pla 
Pa) (ee TRE ey Der 
3 ASD Weta: intl ses LN Ne Ua Sa oD 
(a—b)(6—c) (6—a)(e—a) (a—e)(e—)d) 
i at +f eee lel 
a(a—y)(w@—2) y(y—2z)(y—2) xyz 
a? — be b? +- ac ce + ab 


G@—t(a—e) | @-aG+a C—a)e+s) 


FRACTIONS. 135 


MULTIPLICATION AND DIvIsION uF FRACTIONS. 
165. Multiplication of Fractions. 


The expression : x , means that we are to multiply the 
quotient - by a, and divide the result by 8. 

From the nature of division (§ 76) if we multiply the | 
dividend ¢ by a, we multiply the quotient 5 by a, and 
obtain at if we multiply the divisor d by 0, we divide 
the quotient a by 5, and obtain aa Hence, 

es um 
b bd 

Therefore, to find the product of two fractions, we have 

the following rule: 


Coe 
Aa 


Find the product of the numerators for the required 
numerator, and the product of the denominators for the 
required denominator. 


166, In like manner, 


a 


Cc é ac é ace 
I) rene ees 


ie aracas ii modhs 


and so on for any number of fractions. 


: aV_aya_@ 
Again, eR 52 


In like manner, 


1G TLS, 
(i) = 


167. Division of Fractions. If the product of two numbers 
is equal to 1, each of the numbers is called the reciprocal of 
the other. 


136 SCHOOL ALGEBRA. 


The reciprocal of ; is Z 


a 
b»acrtbe 
for _ -_- 
% ROE as 


The reciprocal of a fraction, therefore, is the fraction 
inverted. 


Since 


and ==]. it follows that 


To divide by a fraction ws the same as to multiply by rts 
reciprocal, 

To divide by a fraction, therefore, 

Invert the divisor and multiply. 


Norr. Every mixed expression should first be reduced to a frac- 
tion, and every integral expression should be written as a fraction 
having 1 for the denominator. If a factor is common to a numera- 
tor and a denominator, it should be cancelled, as the cancelling of a 
common factor before the multiplication is evidently equivalent to 
cancelling it after the multiplication. 
20° x bed, 5.abc_ 

3 ma 5ab- = 8 a’e’d? 
2a7b 6d. 5ab’c 2xX6x5a°id_ 2 


x x = 
3cd? 5ab 8arc?d? 3x5~x 8arbeidt 23 


(1) Find the product of —— 











(2) Find the product of 
C Hiekaciy Lae: 2 si 
Manse able AEP ny RA 
v—say+2y wv+ay («¢#—-y) 
a? — 7? , vy — 2y' ao? — vy 
Re roe a + xy ne 
(x Hy) (e = ay), co 7 “@=y) G atl 
Ws 
ay 


FRACTIONS. 


137 


Norr. The common factors cancelled are (« — y), (1+ y), (c—2y), 


w, and #— y. 








(3) Find the quotient of a a7 age eae 
Orme oO yee, an x (@= 2) (4 + 2) 
(a—2)? a@—2? (a—2)(a— 2) ab 
_ x(a + %). 
b(a— 2) 


The common factors cancelled area and a— x. 


Exercise 50. 


















































Simplify : 
eT laa l4zy) 10. 8a’ be ge ON 
(ie hare a—h ad+tab+ 
9. 3.a7b*c , 20 m'n® Pree Yaa Sere yy 
4amn  2late ey x+y 
3 6@O'?  Smia* 12 ab—b* Oey 
 Imay” Babe  al(a+b) a(a?—b’) 
4. Imin' , 4ary? 13. tase 42" a —2ax 
Saye a a+4axr ax+42° 
a ls 0 ABD, ae ks DE no 
Ol ohn? i Bary Neo eee 
me Tay 3502 15. z* + a zt+3a 
122 36 y2 Hip a SNO Ee tel a 
gy, 42 a — B . a—b 
: cer 16 Shae sa ee 
f PER Ta, Gt d@-ab+ 
Bu? —xX 2a x? — 1 x? — 25 
° oe 17.0 x 
: a 4 22° —Aax v—Ar—-5 x+22—-8 
Si: asda sacar 1S 18. ( =) e a 
52—10 42? ax xy 


138 


19. 


20. 


Pap 


22. 


23. 


24. 


25. 


26. 


27. 


28. 


29. 


30. 


31. 


SCHOOL ALGEBRA. 
pe 
GON NA) 


Creal 42? + 2ary =) 

y° 4¢°4+ 22y+y’ 
ety eo ea eer 

Lary ey? 

BGO ed aa) ica cd 

c ae - cd ae) 

TEENS Pee ere; 
(a+b—e (a+) — & iba Oe 
a+ ab—ac. a Spe b? — be 
ats 0) Oe (Oe ee ty esas) ; 
a—(b—c? e—(a+b) ac—a’?+ab 
(x —a)'— 6") at — (6a), ox + a! —ab 
(7—b—a@ x#—(a—byP be—ab+O 
O20 Ot ey tet Oe 
@+2ab+6—ce a—b-+e 

et et ee a 
x(x +1) v4 2 1 

Qa + 2a°x v—a x eta 


(c—a)(x +a) 2(x iat an 


Gi Ee Bey iad 2¢ ) 
a’ —ab—ac atbte 
taal i AG, a+b Oe De 
a® — b8 a+b A 




















‘ Rabo ee 
39. a+ Tay + ly’, te ey =A 


v+5ryt6y7 2+ d3ay—47 


FRACTIONS. 139 


168. Complex Fractions. A complex fraction is one that 
has a fraction in the numerator, or in the denominator, or 


in both. 
: : 3 
(1) Simplify rast 
4 











eee Wey geal 4 ae | 
4 


Bx Bx _ 32, 4x—1_ 32, 4 





ig 122 
44—-] 


Nore. Generally, the shortest way to simplify a complex fraction 
is to multiply both terms of the fraction by the L.C.D. of the frac- 
tions contained in the numerator and denominator. Thus, in (1), if 

122 
4x—1 





we multiply both terms by 4, we have at once 





Gre a—& 
(2) Simplify “= 2 27%. 
GTe, az 
. a—x# ate 
The L.C. D. of the fractions in the numerator and denom- 
inator is 
(a—x)(a+2). 
Multiply by (a—2)(a+-2), and the result is 


ea ik Cee oe 
(at+a)+(a—2yP 
_ (@ + 2ax + x) — (a — 2ax + 2°) 
Ge Baa 2?) +(e? ar + 24) 
ee as -o — o eae 
6 @ 1 art et Zant 2 
Seed 6 
ei ae ns 
Le eas 
te 








140 SCHOOL ALGEBRA. 


(3) Simphfy a & - 
Lat ae Ae 
He abet x 
l+2+—~_. 42) (lee Pi) a 
l—-“«+2 
‘i x 
yates 
lta«te2 
pie eee iar sind a Ns 
l-Pepe (tse a) 
ote + at | 
1+2 


Nore. In a fraction of this kind, called a continued fraction, we 
begin at the bottom, and reduce step by step. Thus, in the last 


example, we take out the fraction = 


x 


Ne Se ee 
1—27 + 2 


, and multiply 


the numerator and denominator by 1—2 + a?, getting for the result, 


n2. 3 
pound Uo aeck 2) ie i ahy simplified is ea vt + 2" 
(l+a)(l—a#+a*)+¢a l+a423 


Putting this fraction in the given complex fraction for 


4 


’ 


ey eee eke 
l—a#+2? 


we have es ree 
pe oe 

l+e42° 
Multiplying both terms by 1+ 2 + 2°, we get 

a2(1 +a + 23) 
14+@+ 8 >e + a? —28 
oe Ree ery 
1+ 2? 


FRACTIONS. 


Exercise 51. 


Simphify : 























10. 


abe 


12. 


13. 


14. 








zo? 
way ry’ 
a—y 











ab+e @tab 


1 
phate 


afb ea) 
ab 


ey es ee ee, 





1l—2 


141 


142 SCHOOL ALGEBRA. 














15. wet A 
Co corepanyy ot 7 Citarsee nett 
1 Su: 
ee . ae Ts 
Pl 
172 eee iets ee 
a y (ay2 +2 +2) 
Ye 
Z 
Gish ws ane Cre 
3 .abe a b c 
WG Pagan ge tn a ee eee a lel ee 
be +ac+ab J iil 
axe ae 





5 ee ere (Ea ee 


=n on wv 





20. Po Ve S(Goe in ees) 
ety 2#-+y7' z—y #—y¥ 


1 





1 
Pa oie eae ey ; 
éL 1 2be 

a b+e 


21. 





Exercise 52. 
EXAMPLES FOR REVIEW. 
1. Find ‘the value of Va? +6° +e — (a—6b—c)’, when 
Q == 2.6 = 2h and te, 
, 82° + 102+ Te— 2 
2. Reduce to lowest terms Batel eal ae 
Simplify : 
Vi ics A SO ei ms ens Ant Aaa 
(@@—-8)(@—2) (@—1)@—3)' @—)@—2) 


10. 


11. 


12. 


13. 


15. 


16. 


17. 


FRACTIONS. 143 


x + YZ ity y? + x2 2’ + xy 
(Se) a2) (yeaa) (2. eye) 


x Ee) Toner lee \ 
Car aye eee a ) 
Df Lice Bone kL Wisi eee, aman 1a eh 
e\a--e ax+2e xv+cx—2¢ 


ee y' or yf 
xy? E has y? <6 x? aL =} 


OC A) neat peo 


v—x—2 vta—2 x’ —4 
x+y :) ( x ) 
Chap) 

+ 12 4 12 
| ———— 1 2 ree |p 
(1 Sail ei ae) 
oe Me Yi ta Ys (yee) 


etayty e+y (@e+y) 


PAG eee, 20-6 oO Fe a0 





























(ERIN Ce NE BEY CHEN UF Gia BY 
We Lae 2 . AT at yy eae 
zt+l (#+2)(a+1) ie we nae 
Py Sn a eee 
x+2 («+1)(%+2) Gis Nie owen an 








l—2 ,1+22\/2e-1 
j Epes trae ; 
( ian ied Geel) 
—8y\ w@—aty ray). 
R(e— yy et Aay + 4 ey? 
2,2 b+e—a cta—b atb—e 





2 } 
SSE Are Tore ETT : 


Cc be ac ab 


CHAPTER X. 
FRACTIONAL EQUATIONS. 


169. To reduce Equations containing Fractions. 


(1) Solve ease, as) 





3 11 
Multiply by 33, the L.C.M. of the denominators. 
Then, lle —8e +3 = 332 — 297, 
lla —3a—33¢ = — 297 —3; 
— 252 = — 300. 
eral, 


Norte. Since the minus sign precedes the second fraction, in remov- 
ing the denominator, the + (understood) before a, the first term of the 
numerator, is changed to —, and the — before 1, the second term of 
the numerator, is changed to +. 


Therefore, to clear an equation of fractions, 


Multiply each term by the L.C.M. of the denominators. 


If a fraction is preceded by a minus sign, the sign of 
every term of the numerator must be changed when the 
denominator 1s removed. 








—~— A Be ae eA oo 8 
2 lve & yA ee cae Ae 
(2) Soa ate ea—6. «£—-8- 2-—9 


Norr. The solution of this and similar problems will be much 
easier by combining the fractions on the left side and the fractions 
on the right side than by the rule given above. 


(x — 4) (x — 6) —(w— 5? _ (w@ —7) (w@ — 9) — (a — 87 
(e—5)(a—6) (x — 8) (% — 9) 


FRACTIONAL EQUATIONS. 145 


By simplifying the numerators, we have 


sede 5° I ae ee 
(7—5)(e@—6) (x—8)(e%—9) 


Since the numerators are equal, the denominators are equal. 
Hence, (x — 5) (x — 6) = (w — 8) (w — 9). 


Solving, we have a = 7. 


Exercise 53. 





























Solve: 
se Ofimahon ate be PAG ero 
4 3 5 
piled i Goel ae 197+3 
6 4 
= Setl, Wert 3e-1_ je) 
3B 9 a 6 
Ome ee bla oe 
Em ee eee ere te he 
TA 2 14 8 
5 Ohi A Boe DT Ae | 
3 
Scr a ee ee a 
cometG ney 8 
7 Bie ey Bel, BO ate ane 
ai 14, 56 56 
BO Mie Ode 1 
= , 3 12 v4 
382—-1 , 2%+1 227—5., Tx—1 
9. Lb] | ——— + | = 10 |, + 
( 4 a5 3 ) ( 3 : 8 j 
Tza—-4 ,8a—1 5(x4—1) 3882-1), 2 
10. AUESI Soe MGS OSES ae ele ID 
9g a eS 6 20 Ty 


146 


ri 


La. 


13. 


14. 


15. 


16. 


Wh Ue A 


18. 


19. 


24. 


25. 


26. 


27. 


28. 


SCHOOL ALGEBRA. 


ae aaa) nd Baar inch ie id 
































62 ee ee eee 22, 
ef 4 5 4 avail 
10z+11_ 122-13, 7-62 

6 3 4 

3 5 9 1 
Wf 4 ee 
aed M4 CIT mee EP ACOE eke) 1 
S20 bas Ow ak Omg Ly eo 
x—l 2(«-1) 38(x—-1) w-—1 18 
22—8 g= 25 a0 
Qa —4 8r—6 2 
Le (ne ae 20 22-1 8 a 2a 
6-—Tx 5a er RD ike 6 eee a Rl te 3, 
B+82_45--82) ,, 5-22 2—Te_ Sat+4 
Set Oona Gee ae te ee Rae er Wor rare, 
Omi LE OO ere Mens Gp eae! a ea 
Das ope 8 Ciel D0) pe EON Miers et Se 
Ga) eee 4 peri A ah oie 








Ze 
Wa its rotons (ree | ltw l—a« 1-2 
2a+l.T7¢a#—1 22?—8x2—45 


oo — 


8x2—8 62+6 42’°— 4 
V’—-a“+tl, #xtetl 














a Se 
x—l i zc+l1 zs 
92+5 4 Bal ble Sih eet 
6(¢—1) Iota) 9(#+1) 
Ace eae ot pie 
2419 ¢+8 2 ba+6 
1 1 1 1 














FRACTIONAL EQUATIONS. 147 


170. If the denominators contain both simple and com- 
pound expressions, it is best to remove the simple expressions 
first, and then each compound expression in turn. After 
each multiplication the result should be reduced to the 
simplest form. 


8e+5, fe—38_ 4276 
14 62+2 7 

Multiply both sides by 14. 

rent 80 +54 We) 80412, 


Transpose and combine, IG e=9) <7. 
3a + 


Divide by 7 and multiply by 32 +1, 
Te—3=32+41. 


w=. 








(1) Solve 








(oiccive ela 
BC AE 10 


Simplify the complex fractions by multiplying both terms of each ) 
fraction by 9. 
Then, 27—42 1 Ta — 27 


Multiply both sides by 180. 
135 — 20a = 45 — 14x + 54, 
—6a=— 36. 
“0 = 6. 


Exercise 54. 
Solve: 


La eats reg) ae OE Semel 
10 ee arta 


de 20_¢2— 12, & 
36 ba—4 4 





148 SCHOOL ALGEBRA. 
Ogi awe 2) f be 
18 1382—16 9 


4 Grrl whi 182 oes 
9 62+8 3 
































5. mectipas NedetRS Yaa AES Ca ak 
15 Tz—6 5 
6 Era Needing Eola 
es 7x —16 5 
7 Lig me LG ee eh Ce 
14 28 2(82-+7) 
a 2¢4- 8h) 13a — 2 era eta LO: 
9 Lice separa 36 
9) Baba.) Mie een Mel Ori amen 2h 
1S a ee) ee 6 105 
2a—5 to 4x%—8 
10. =s — lL. 
5 2x4 —15 10 ls 


171. Literal Equations. Literal equations are equations 
in which some or all of the given numbers are represented 
by letters; the numbers regarded as known numbers are 
usually represented by the first letters of the alphabet. 

(1) (a—2)(a+2)= 20°+ Q2axr— 2". 

Then, a? — x? = 207 + 2ax— 2}, 


— 2ax =a’, 


Exercise 55. 
Solve: 


1. az~+2b6=3bx+4a. 3. (a+2)(6+2)=2(2 - c). 
2. 27+60=(a—2V. 4. (x—a)(x+ 6)=(x—b)(x—c). 


Lb. i= 


16. 


FRACTIONAL EQUATIONS. 149 



















































































RMS 17, 24% 22\— a" 
atax e+cx jtenee,) aes 
c+d _ mx Tee a oe DY 
ab+bz an+nz Mukono 
oo aa m+n 
re 
M+n_M—N iA a, Ping 
2+2 2-2 m 
a+be e+dxr ax—b axrt+b 
a+b e+d a Py ie 
6a+a_84—b Mig 6 
4r+b6 2x-a b a 
Lier ca +d 
, —--—tL-=d, 
aa a a _ 24d 
Cx a 
EE gin eons Be 2 
c a 
2G ey aa, ape—3 
OS PAE 5-5 
eM Oe ih, 2ba Z : ‘i 
a—b a+b w—6? 
1 n «“2+_n 23. ire Oar Oct Dike 
a Saas Bia de bate ax—e 
a aie ue a an tta «—b__2(a+b) 
a b—a b+a x—a x«x+b 2 
xta zt—b,c—bzx 
25. =b — 
b C v b 
20a—bx , 9e—ax , 6d—cx 
26. 10), 
5a ne 3¢ ay 2d . 
ax b—xz , a(b—x 
27. —— =eih, 
b 2c a 3d i 


150 SCHOOL ALGEBRA. 


172. Problems involving Fractional Equations. 


Ex. The sum of the third and fourth parts of a certain 
number exceeds 3 times the difference of the fifth and sixth 
parts by 29. Find the number. 


Let x =the number. 
Then at ri the sum of its third and fourth parts, 
ae a the difference of its fifth and sixth parts, 


6 


er ae 3 times the difference of its fifth and sixth parts, 
; “ : —3 G — 4 = the given excess. 
But 29 = the given excess. 
BH-s(E-f)-» 
Multiply by 60 the L.C. D. of the fractions. 
20% + 152—36a + 30x = 60 x 29, 
Combining, 29 x = 60 x 29. 
*. «= 60, 


Exercise 56. 


1. The sum of the sixth and seventh parts of a number 
is 13. Find the number. 


2. The sum of the third, fourth, and sixth parts of a 
number is 18. Find the number. 


3. The difference between the third and fifth parts of 
a numbef is 4. Find the number. 


4. The sum of the third, fourth, and fifth parts of a 
number exceeds the half of the number by 17. Find the 


number. 


5. There are two consecutive numbers, z and z+1, 
such that one-fourth of the smaller exceeds one-ninth of the 
larger by 11. Find the numbers. 


* 
FRACTIONAL EQUATIONS. 151 


6. Find three consecutive numbers such that if they are 
divided by 7, 10, and 17, respectively, the sum of the quo- 
tients will be 15. 


7. Find a number such that the sum of its sixth and 
ninth parts shall exceed the difference of its ninth and 
twelfth parts by 9. | 


8. The sum of two numbers is 91, and if the greater is 
divided by the less the quotient is 2, and the remainder 
is]. Find the numbers. 


Dividend — Remainder 


HInt. _= 
Divisor 


= Quotient. 

9. The difference of two numbers is 40, and if the greater 
is divided by the less the quotient is 4, and the remainder 
4. Find the numbers. 


10. Divide the number 240 into two parts such that the 
smaller part is contained in the larger part 5 times, with a 
remainder of 6. 


11. Ina mixture of alcohol and water the alcohol was 
24 gallons more than half the mixture, and the water was 
4 gallons Jess than one-fourth the mixture, How many 
gallons were there of each ? 


12. The width of a room is three-fourths of its length. 
If the width was 4 feet more and the length 4 feet less, the 
room would be square. Find its dimensions. 


Ex. Eight years ago a boy was one-fourth as old as he 
will be one year hence. How old is he now? 


Let x = the number of years old he is now. 
Then x— 8 =the number of years old he was eight years ago, 
and xz +1 =the number of years old he will be one year hence. 


“ £—-8=43(x +1). 
Solving, x=11. 


152 SCHOOL ALGEBRA. 


13. A is 60 years old, and B is three-fourths as old. 
How many years since B was just one-half as old as A? 


14. A father is 50 years old, and his son is half that 
age. How many years ago was the father two and one- 
fourth times as old as his son? 


15. A father is 40 years old, and his son is one-third 
that age. In how many years will the son be half as old 
as his father ? 


Ex. A can do a piece of work in 6 days, and B can do 
it in 7 days. How long will it take both together to do 
the work ? 


Let « = the number of days it will take both together. 
Then +-=the part both together can do in one day, 
Ae 
% = the part A can do in one day, 
+= the part B can do in one day, 
and i-+2= the part both together can do in one day. 
Lyla 
~sy ed 
5 Wid si 
Ta + 6a = 42, 
13 @ = 42, 
t= 3735. 


Therefore they together can do the work in 3,8 days. 


16. A can do a piece of work in 8 days, B in 4 days, and 
Cin 6 days. How long will it take them to do it working 
together ? 


17. A can do a piece of work in 24 days, B in 34 days, 
and C in 42 days. How long will it take them to do it 
working together ? 


18. A and B°can separately do a piece of work in 12 
days and 20 days, and with the help of C they can do it in 
6 days. How long would it take C to do the work? 


FRACTIONAL EQUATIONS. 153 


19. A and B together can do a piece of work in 10 days, 
A and C in 12 days, and A by himself in 18 days. How 
many days will it take B and C together to do the work ? 
How many days will it take A, B, and C together ? 


20. A and B can do a piece of work in 10 days, A and 
C in 12 days, B and C in 15 days. How long will it take 
them to do the work if they all work together ? 


21. A cistern can be filled by three pipes in 15, 20, and 
30 minutes respectively. In what time will it be filled if 
the pipes are all running together? 


22. A cistern can be filled by two pipes in 25 minutes 
and 30 minutes, respectively, and emptied by a third in 20 
minutes. In what time will it be filled if all three pipes 
are running together ? 


23. A tank can be filled by three pipes in 1 hour and 20 
minutes, 2 hours and 20 minutes, and 8 hours and 20 min- 
utes, respectively. In how many minutes can it be filled 
by all three together ? 


Ex. A courier who travels 6 miles an hour is followed, 
after 5 hours, by a second courier who travels 74 miles an 
hour. In how many hours will the second courier overtake 
the first ? 


Let x = the number of hours the first travels. 
Then x — 5 =the number of hours the second travels, 
6a =the number of miles the first travels, 
and (~ — 5) 74 = the number of miles the second travels. 
They both travel the same distance. 
62 = (w — 5) 74, 
or 12% = 15% — 75. 
ie eee 20, 


Therefore the second courier will overtake the first in 20 
hours. 


154 SCHOOL ALGEBRA. 


24. A sets out and travels at the rate of 9 miles in 2 
hours. Seven hours afterwards B sets out from the same 
place aud travels in the same direction at the rate of 5 
miles an hour. In how many hours will B overtake A? 


25. A man walks to the top of a mountain at the rate of 
23 miles an hour, and down the same way at the rate of 4 
miles an hour, and is out 5 hours. How far is it to the top 
of the mountain ? 

26. In going from Boston to Portland, a passenger train, 
at 27 miles an hour, occupies 2 hours less time than a freight 
train at 18 miles an hour. Find the distance from Boston 
to Portland. 

27. A person has 6 hours at his disposal. How far may 
he ride at 6 miles an hour so as to return in time, walking 
back at the rate of 3 miles an hour? ° 

28. A boy starts from Exeter and walks towards Ando- 
ver at the rate of 3 miles an liour, and 2 hours later another 
boys starts from Andover and walks towards Exeter at the 
rate of 24 miles an hour. The distance from Exeter to 
Andover is 28 miles. How far from Exeter will they meet? 


Ex. A hare takes 5 leaps while a greyhound takes 3, but 
1 of the greyhound’s leaps is equal to 2 of the hare’s. The 
hare has a start of 50 of her own leaps. How many leaps 
must the greyhound take to catch her? 

Let 3a =the number of leaps the greyhound takes. 

Then 5x2=the number of leaps the hare takes in the same time. © 

Also, let a =the number of feet in one leap of the hare. 

Then 2a=the number of feet in one leap of the hound. 

Hence 32x 2a or 6ax =the whole distance, 
and (5e+50)a or 5ax + 50 = the whole distance. 

. 6ax = 5axz + 50a, 
x= 50, 

and : 3¢'= 150. 

Therefore the greyhound must take 150 leaps. 


FRACTIONAL EQUATIONS. 155 


29. A hare takes 7 leaps while a dog takes 5, and 5 of 
the dog’s leaps are equal to 8 of the hare’s. The hare has 
a start of 50 of her own leaps. How many leaps will the 
hare take before she is caught ? 


30. A dog makes 4 leaps while a hare makes 5, but 8 of 
the dog’s leaps are equal to 4 of the hare’s. The hare has a 
start of 60 of the dog’s leaps.. How many leaps will each 
take before the hare is caught? 


Norse. If the number of units in the breadth and length of a 
rectangle are represented by x and x +a, respectively, then x(x + a) 
will represent the number of units of area in the rectangle. 


31. A rectangle whose length is 24 times its breadth 
would have its area increased by 60 square feet if its length 
and breadth were each 5 feet more. Find its dimensions. 


32. A rectangle has its length 4 feet longer and its width 
3 feet shorter than the side of the equivalent square. Find 
its area. 


33. The width of a rectangle is an inch more than half 
its length, and if a strip 5 inches wide is taken off from the 
four sides, the area of the strip is 510 square inches. Find 
the dimensions of the rectangle. 


Note. If « pounds of metal lose 1 pound when weighed in 


water, 1 pound of metal will lose l ofa pound. 
ce 


34. If 1 pound of tin loses 4% of a pound, and 1 pound 
of lead loses 3% of a pound, when weighed in water, how 
many pounds of tin and of lead in a mass of 60 pounds that 
loses 7 pounds when weighed in water ? 


35. If 19 ounces of gold lose 1 ounce, and 10 ounces of 
silver lose 1 ounce, when weighed in water, how many 
ounces of gold and of silver in a mass of gold and silver 
weighing 530 ounces in air and 495 ounces in water ? 


156 SCHOOL ALGEBRA. 


Ex. Find the time between 2 and 8 o'clock when the 
hands of a clock are together. 


At 2 o’clock the hour-hand is 10 minute-spaces ahead of the 
minute-hand. 
Let ~ «=the number of spaces the minute-hand moves over. 
Then x—10= the number of spaces the hour-hand moves over. 
Now, as the minute-hand moves 12 times as fast as the hour-hand, 
12(2 — 10) =the number of spaces the minute-hand moves over. 


*. w= 12(%—10), 
and 11a = 120. 
*, = 1029, 


Therefore the time is 101% minutes past 2 o'clock. 


36. At what time between 2 and 3 o'clock are the hands 
of a watch at right angles? 


37. At what time between 3 and 4 o’clock are the hands 
of a watch pointing in opposite directions ? 


38. At what time between 7 and 8 o’clock are the hands 
of a watch together? 


Ex. A merchant adds yearly to his capital one-third of 
it, but takes from it, at the end of each year, $5000 for 
expenses. At the end of the third year, after deducting 
the last $5000, he has twice his original capital. How 
much had he at first? 


Let « = number of dollars he had at first. 

Then <2 _ 5000, FA Ne cll 2a) 
will stand for the number of dollars at the end of first year, 
nee : (ee aa 2 S00 or 16x tat 


will stand for the number of dollars at the end of second year, 
Ay a) Ce ~ aa? 25000, "or 64% — 555000 
3 9 27 


will stand for the number of dollars at the end of third year. 


FRACTIONAL EQUATIONS. 157 


But 22 stands for the number of dollars at the end of third year. 
642 — 555000 _ 
ms Brel a ee 

Whence x = 55,500. 


20. 


39. A trader adds yearly to his capital one-fourth of it, 
but takes from it, at the end of each year, $800 for ex- 
penses. At the end of the third year, after deducting the 
last $800, he has 14 times his original capital. How much 
had he at first? 


40. A trader adds yearly to his capital one-fifth of it, 
but takes from it, at the end of each year, $2500 for ex- 
penses. At the end of the third year, after deducting 
the last $2500, he has 1,5 times his original capital. 
Find his original capital. 


41. A’s age now is two-fifths of B’s. Eight years ago A’s 
age was two-ninths of B’s. Find their ages. 


42. A had five times as much money as B. He gave B 
5 dollars, and then had only twice as much as B. How 
much had each at first ? 


43. At what time between 12 and 1 o'clock are the hour 
and minute hands pointing in opposite directions ? 


44. Hleven-sixteenths of a certain principal was at in- 
terest at 5 per cent, and the balance at 4 per cent. The 
entire income was $1500. Find the principal. 


45. A train which travels 36 miles an hour is # of an 
hour in advance of a second train which travels 42 miles 
an hour. In how long a time will the last train overtake 
the first ? 


46. An express train which travels 40 miles an hour 
starts from a certain place 50 minutes after a freight train, 
and overtakes the freight train in 2 hours and 5 minutes. 
Find the rate per hour of the freight train. 


158 SCHOOL ALGEBRA. 


47. A messenger starts to carry a despatch, and 5 hours 
after a second messenger sets out to overtake the first in 8 
hours. In order to do this, he is obliged to travel 24 miles 
an hour more than the first. How many miles an hour 
does the first travel ? 


48. The fore and hind wheels of a carriage are respec- 
tively 94 feet and 112 feet in circumference. What distance 
will the carriage have made when one of the fore wheels 
has made 160 revolutions more than one of the hind 
wheels ? 


49. When a certain brigade of troops is formed in a 
solid square there is found to be 100 men over; but when 
formed in column with 5 men more in front and 3 men less 
in depth than before, the column needs 5 men to complete 
it. Find the number of troops. 


50. An officer can form his men in a hollow square 14 
deep. The whole number of men is 8136. Find the num- 
ber of men in the front of the hollow square. 


51. A trader increases his capital each year by one- 
fourth of it, and at the end of each year takes out $2400 
for expenses. At the end of 3 years, after deducting the 
last $2400, he finds his capital to be $10,000. Find his 


original capital. 


52. A and B together can do a piece of work in 14 days, 
A and C together in 1% days, and B and C together in 1% 
days. How many days will it take each alone to do the 
work ? | 


53. A fox pursued by a hound has a start of 100 of her 
leaps. The fox makes 3 leaps while the hound makes 2; 
but 3 leaps of the hound are equivalent to 5 of the fox. 
How many leaps will each take before the hound catches 
the fox? 


FRACTIONAL EQUATIONS. 159 


173, Formulas and Rules. When the given numbers of a 
problem are represented by letters, the result obtained from 
solving the problem is a general expression which includes 
all problems of that class. Such an expression is called a 
formula, and the translation of this formula into words is 
called a rule. 


We will illustrate by examples : 


(1) The sum of two numbers is s, and their difference d; 
find the numbers. 














Let * g =the smaller number; 
then x + d =the larger number. 
Hence etet+d=s, 
or 20 = §.— d. 
eee te d, 
2 
and nied @ ones 
2 
se 
2 
Therefore the numbers are ° . d and © = d. 


As these formulas hold true whatever numbers s and d 
stand for, we have the general rule for finding two numbers 
when their sum and difference are given: 


Add the difference to the sum and take half the result for 
the greater number. 

Subtract the difference from the sum and take half the 
— result for the smaller number. 


(2) If A can do a piece of work in q@ days, and B can 
du the same work in 6 days, in how many days can both 
together do it? 


160 SCHOOL ALGEBRA. 





Let a =the required number of days. 
Then, 1 _ the part both together can do in one day. 
x 
Now 1 _ the part A can do in one day, 
a 
and += the part B can do in one day ; 
Bee her can do in one da 
therefore) =—+ ve the part both together can do in one day. 
a 
ead 
an one 
Wh oe 
ence aT 


The translation of this formula gives the following rule 
for finding the time required by two agents together to 
produce a given result, when the time required by each 
agent separately is known : 


Dude the product of the numbers which express the units 
of tume required by each to do the work by the sum of these 
numbers ; the quotrent is the tume required by both together. 


174. Interest Formulas. The elements involved in com- 
putation of interest are the principal, rate, time, mterest, 
and amount. 

Let y= the principal, 

y = the interest of $1 for 1 year, at the given rate, 
¢ = the time expressed in years, 

2 = the interest for the given time and rate, 

a = the amount (sum of principal and interest). 


175. Given the Principal, Rate, and Time; to find the Interest. 


Since 7 is the interest of $1 for 1 year, pr is the interest 
of $p for 1 year, and prt is the interest of $p for ¢ years. 


ehh jet | (Formula 1.) 


Rue. Multiply together the principal, rate, and time. 


FRACTIONAL EQUATIONS. 161 


176. Given the Principal, Rate, and Time; to find the Amount. 
Since the amount a is the sum of the principal and 
interest, 
a—p-t prt. (Formula 2.) 
177. Given the Amount, Rate, and Time; to find the Principal. 
From formula 2, p+pri=a, 
or pil-- rt) =a. 
Divide by 1+ rt, Deas (Formula 3.) 


178. Given the Amount, Principal, and Rate; to find the Time, 
From formula 2, pt pri=a. 
Transpose 7, pri=a— p. 


Divide by pr, t= aie (Formula 4.) 


179. Given the Amount, Principal, and Time; to find the Rate. 
From formula 2, p+prt=a. 
Transpose 7, pri=a— p. 
Divide by pt, te ec (Formula 5.) 
Exercise 57. 
Solve by the preceding formulas: 


1. The sum of two numbers is 40, and their difference is 
10. Find the numbers. 


2. The sum of two angles is 100°, and their difference is 
21° 30’. Find the angles. 

8. The sum of two angles is 116° 24’ 80", and their 
difference is 56° 21' 44". Find the angles. 


162 SCHOOL ALGEBRA. 


4. A can do a piece of work in 6 days, and B in 5 days. 
How long will it take both together to do it? 


5. Find the interest of $2750 for 3 years at 4% per 
cent. 

6. Find the interest of $950 for 2 years 6 months at 5 
per cent. 

7. Find the amount of $2000 for 7 years 4 months 
at 6 per cent. 

8. Find the rate if the interest on $680 for 7 months is 
$395.70. 


9. Find the rate if the amount of $750 for 4 years is 

$900. 

10. Find the rate if a sum of money doubles in 16 years 
and 8 months. 

11. Find the time required for the interest on $2130 to 
be $436.65 at 6 per cent. 

12. Find the time required for the interest on a sum of 
money to be equal to the principal at 5 per cent. 

13. Find the principal that will produce $161.25 interest 


in 3 years 9 months at 8 per cent. 


14. Find the principal that will amount to $1500 in 3 
years 4 months at 6 per cent. 


15. How much money is required to yield $ 2000 interest 
annually if the money is invested at 5 per cent? 


16. Find the time in which $640 wll amount to $1000 
at 6 per cent. 

17. Find the principal that will produce $100 per month, 
at 6 per cen!, 


18. Find the rate if the interest on $700 for 10 months 
is $25, 


CHAPTER XI. 


SIMULTANEOUS EQUATIONS OF THE FIRST 
DEGREE. 


180. If we have two unknown numbers and but one rela- 
tion between them, we can find an unlimited number of 
pairs of values for which the given relation will hold true. 
Thus, if z and y are unknown, and we have given only the 
one relation x-+y=10, we can asswme any value for 2, 
and then from the relation 2+ y= 10 find the correspond- 
ing value of y. For from x-+y=10 we find y=10—z. 
If stands for 1, y stands for 9; if x stands for 2, y stands 
for 8; if # stands for ~ 2, y stands for 12; and so on with- 
out end. 


181, We may, however, have two equations that express 
different relations between the two unknowns. Such equa- 
tions are called independent equations. Thus, «+ y= 10 
and «—y—=2 are independent equations, for they evidently 
express different relations between x and y. 


182. Independent equations involving the same unknowns 
are called simultaneous equations. 

If we have two unknowns, and have given two independ- 
ent equations involving them, there is but one pair of values 
which will hold true for both equations. Thus, if in § 181, 
besides the relation 2+ y = 10, we have also the relation 
x — y = 2, the only pair of values for which both equations 
will hold true is the pair x= 6, y= 4. 

Observe that in this problem 2 stands for the same num- 
ber in both equations; so also does y. 


164 SCHOOL ALGEBRA. ~ 


183. Simultaneous equations are solved by combining 
the equations so as to obtain a single equation with one 
unknown number; this process is called elimination. 

There are three methods of elimination in general use: 


I. By Addition or Subtraction. 
If. By Substitution. 
III. By Comparison. 


184, Elimination by Addition or Subtraction. 


(1) Solve: 54 — 38y= 20 (1) 
22+ 5y = 39 (2) 
Multiply (1) by 5, and (2) by 3, 
| 25 —15y = 100 (3) 
62+ 15y=117 (4) 
Add (3) and (4), 3la = 217 
hala 
Substitute the value of x in (2), 
14 + 5y = 39, 
“» ¥ =); 


In this solution y is eliminated by addtvtion. 








(2) Solve: 62+ 385y=177 \ (1) 
82—2ly= 331 (2) 
Multiply (1) by 4, and (2) by 3, 
24x + 140y = 708 (3) 
24x%— 63y= 99 (4) 
Subtract, 203 y = 609 
. ¥ =3. 
Substitute the value of y in (2), 
8x2 — 63 = 33. 
“. = 12. 


In this solution 2 is eliminated by subtraction. 


SIMULTANEOUS EQUATIONS. 165 


185. Hence, to eliminate by addition or subtraction, we 
have the following rule: 


Multiply the equations by such numbers as will make the 
coefficients of one of the unknown numbers equal in the 
resulting equations. 

Add the resulting equations, or subtract one from the other, 
according as these equal coefficients have unleke or like signs. 

Note. It is generally best to select the letter to be eliminated 
which requires the smallest multipliers to make its coefficients equal ; 
and the smallest multiplier for each equation is found by dividing 
the L.C.M. of the coefficients of this letter by the given coefficient in 
that equation. Thus, in example (2), the L.C.M. of 6 and 8 (the co- 


efficients of x) is 24, and hence the smallest multipliers of the two 
equations are 4 and 3 respectively. 


Sometimes the solution is simplified by first adding the 
given equations, or by subtracting one from the other. 


(3) a+49y= 51 (1) 
49n+ y= 99 (2) 
Add (1) and (2), 50a + 50y = 150 (3) 
Divide (3) by 50, c+y=3., (4) 
Subtract (4) from (1), 48 y = 48. 
y=, 
Subtract (4) from (2), 48 x = 96. 
2s 


Exercise 58. 


Solve by addition or subtraction : 


ule Pamaaae | 4. SO ae 
27—- y= 3 o2—6y=15 
2. ibaa | 5. Saar pal 
2a—4y= 4 382—Q2y=17 
3. cade 6. RU Gast 
z+ Sy = 28 22—d3y = 26 


166 SCHOOL ALGEBRA. 


Ts cab | pe ee hd ca 3 
Deen en az2— 3y= 90 
8. Sten ace 12. °47— amine! 
22+38y = 48 842— 4y=17 
9. oes 13. Tx— eee 
5a —2y=15 x —10y= 39 
10. 22+ y= a 14. 84+ aes 
Ta+5y=21 22+ 5y=13 


186, Elimination by Substitution. 


(1) Solve: idence | 
4¢+3y=25 
Ba +4y=32 (1) 
4a + 3y =25 (2) 
Transpose 4y in (1), 5a=32—4y. (3) 
Divide by coefficient of 2, Apia abe s (4) 


Substitute the value of & in (2), 
4 Gen ) 434 = 25, 


5 
128 : 16Y 4 34 —95, 


128 — 16y + 1dy = 125, 
—y=—3. 
“. Y= ds 
Substitute the value of y in (2), 
4x4 +49 = 25. 
*. B= 4, 


Hence, to eliminate by substitution, 


From one of the equations obtain the value of one of the 
unknown numbers in terms of the other. 

Substitute for this unknown number its value in the other 
equation, and reduce the resulting equation. 


Solve by substitution : 


SIMULTANEOUS EQUATIONS. 


Exercise 59. 


167 


tee ra 8. 3x%— amet 
De BD Yani | 22+ d5y= 63 
Ae } 9. 2x — ute 
of¢— 2y=10 o@-+ 2y = 29 
8. 227— 3y=> a 10. 62 — ape 
of 2y= 29 Da— by=, 8 
4, x Leese 11. fet ee 
20+ Ty=88 22+ 5y = 382 
Ge ee a 12a ELS 
x+ 2y=25 oa+ 2y = 46 
6. Sat oh ae, 13. 382— Aa 
l3e2— dy=21 4n— dy = 9 
ie vs 14. 54+ 9y=—1 
(te A al 8a+1lly= 
187. Elimination by Comparison. 
Solve: Stee a rth 
342+ 2y = 23 
22 —dy = 66. 
3x + Qy = 23. 
Transpose 5y in (1), and 2y in (2), 
Qe = 66 +5y, 
3% = 23 —2y. 
Divide (3) by 2, iz me 
Divide (4) by 3, ne ee 


Equate the values of z, 


66+ 5y _ 


23 — 2y 
3 


168 SCHOOL ALGEBRA. 


Reduce (7), 198 + 15y = 46 —4y, 
19y = — 152. 
“ y¥=—8., 
Substitute the value of y in (1), 
2x + 40 = 66. 
*. w= 13. 


188. Hence, to eliminate by comparison, 


From each equation obtain the value of one of the unknown 
numbers in terms of the other. 

Form an equation from these equal values and reduce the 
equation. 


Exercise 60. 


Solve by comparison : 


1 86+) 6 6y=30 


382— 2y= 25 Sa+ 2y=126 
2. Tea+ By= 70 109000 9 = eel 
5a— 4y= 7 ie eat) 
3. 9xe-+ 4y=—54 11 24+2Zly= 2 
4xr+ 9y=89 aly+ 2e= 19 
82-4 y= 8 382+ 10y=125 
5. 2x —38y= 29 13. 62—l18y= 2 


5a—l2y= 4 


14. 2r+ .y=108 
102+ 2y= 60 


15. 382— 5By=— 5 


16. 122-- Ty=176 
8y—19727= 8 


; ee 
4. Tx+ eae 12. 10x+ aa 
) 


18z—2y= 57 


SIMULTANEOUS EQUATIONS 169 


189, Each equation must be simplified, if necessary, 
before the elimination. 


Solve: Se—tyt halt 
1@+)+8y-)=9 
go 2 Yh tel. (1) 
g@t+1)+ey—H=9. (2) 
Multiply (1) by 4, and (2) by 12, 

34 —2y—2=4, (3) 
de +449y—9= 168, (4) 
From (3), 3a—2y =6. (5) 
From (4), 4¢4+9y=1138. ; (6) 


Multiply (5) by 4, and (6) by 3, 
122— 8y= 24 
12% 4+ 27y = 339 
35y = 315 
oo af eee 
Substitute value of y in (1), | t= 8! 


Exercise 61. 














Solve 
Co Y aw e—l) y— 2 
1 242 Se 7 Weis ae Aes ay 
309 2 | Sy 
cyl 9 BY a Ul 
973 | 53 J 
Ea Ue BGs oe2—dYy Zu + yf | 
Pee 4 = 9 5 Fs] 
Beers D | es 5 | 
sec ea is pares gta ay _ 2 y 
gt 9 | a 513 | 
3 ee Lee gt sts 3Y _¢ 
40 is 
2a—Y 1 9 ued | (2 el ayy | 
Ss edge J 








12. 


13.) 


14. 


SCHOOL ALGEBRA. 


9 gee 
30 


2—2 10—az _-y—10 


Sn eee 
5 — 


4 


ay + Ay eae 26 
3 Shore te nen 


8 














2a yt Sen —2y 713 
% 4 





2y+1 


6x—5y+4_ 382+ 
3 


19 


3 x+2 jel 5 
oe 

yy ela 
Hie) os 


ee. 
ee et ee iN a ee 


| 


| 


SIMULTANEOUS EQUATIONS. 171 





15. ety __9 TG: pene Seer el 
ee | 4 10 

Us ake | &,yt2_ 91 

fi 6 ; 


afl L te l0g Sy 20. 32-24-43 
3 20 30 


Nore. In solving the following problems proceed as in 3 171. 
19. 6y +5 saa (oh inst oS *| 





8 4¢4—2y 
Bay 8s tee yaa Otis 
4 7—2 3 


Dap 
20. Rosnayaca tt 
Pee rome 52 
Uiqleme ye y 3 


at fete? me Sy EE Se 


2x2+8 6 
Ogata te Oy toe lO a 58 
4 22 —3y 8 
22 RE OLRNO Ce = ay) e ol lae ae 
SitetO 2(a — 4) 5 
3(2y+3)_ 6y+21  3y+5z 
4 4 2(2y— 3) 


172 SCHOOL ALGEBRA. 


190, Literal Simultaneous Equations. 


Solve: ‘ax + by Sc ; 
ae bly 

Nort. The letters a’, b’ are read a prime, b prime. In like man- 
ner, a/’, a/’’ are read a second, a third, and ay, da, dg are read @ sub 
one, a sub two, a sub three. It is sometimes convenient to represent 
different numbers that have a common property by the same letter 
marked by accents or suffives. Here a and a’ have a common 
property as coefficients of a. 


az +by =c. (1) 
a/c + b’/y=c’, (2) 
To find the value of y, multiply (1) by a’, and (2) by a, 
aa’x + a/by =a/ec 
aa’e + ab/y = ac’ 
a/by — ab’/y = a/c — ac’ 
a/c — ac’ 
a/b — ab’ 
To find the value of x, multiply (1) by 07, and (2) by 4, 
ab/a + bb’y = b/c 
a/be + bb’y = be’ 
ab/x — a/bx = b’c — be’ 
__ b/c — be’ 
ab/—a’b 


Us 





Exercise 62. 


Solve: 

7 Piva 5. gas oi 
r—-y=a x= dy 

2. nat + nly ae 6-1 Oe ea 
Ws + n'y = yr be — aly 1 

3. ane :- ioe ae ee 
ate + bly =e! 4bx— 8ay = tab 

4. x£2— ace 8. 2S Bean ees 
cx + aby = ms 82 —-2y=a+b 


SIMULTANEOUS EQUATIONS. 

















173 




















ba Tee i 
: vienna FAN 
bz +cy=a+b 
xr+m_a ‘14. eee 
10. = = 
y—n 5b bz +ay=c 
ba a 
ae y 15. 38a? + Seige Raat 
116 et EN rey ax+-2by=d > 
eneD 
2 y_} 1G a ew ble) 
a bles a+b a—b a+b | 
E 
py, ae pets.) Chi vik ot ane 
a cl a SG, a—b } 
e+y=2a 
righ ee nteN i an 
13. Ze, 4y_ Sy 2 m—-a m—b 
3a sno 6 b a . ) 
pie) Ga 
t—y=a—b payee ae, 


191. Fractional simultaneous equations, of which the de- 


nominators are simple expressions and contain the unknown 
numbers, may be solved as follows: 


(1) Solve: 


We have 


and 
To find the value of y. 
Multiply (1) by ¢, 


ab | 
= —-=™m 
ey 
Cee | 
-+-=n 
cy J 
i dee (1) 
way 
a (2) 
zo y| 
Ad yk, (3) 
eames A 


we! SCHOOL ALGEBRA. 


Multiply (2) by a, mnGinesaa 5, 


Subtract (4) from (3), 





Multiply both sides by y, be —ad=(cm — an) y. 











_ bc—ad 
~ em—an 
To find the value of «x. 
ad bd 
Multiply (1) by d, — +—=dm. 
a ae 
Multiply (2) by 8, 2g + pao bn, 
ei. 
Subtract (6) from (5), CLs ee bn. 
z 
Multiply both sides by x, ad — be = (dm — bn)a. 
. _ ad — be. 
dm — bn 
(2) Solve: 5 2g 
oa dy 
ae 4 
av de OEY, 
erg aes 
Weh says RES 
e have Sick ay 7 
and Ble 5/7 oe 3. 
62 l0y 
Multiply (1) by 15, the L.C.M. of 3 and 5, and (2) by 30, 
a + oe 105 
be 
sad Sas 
TOR 
Multiply (4) by 2, and add the result to (3), 
9 285, 
x 
1 
to 
3 


Substitute the value of x in (1), and we get 


eet, 
are 


(4) 


(5) 
(8) 


(2) 


(3) 
(4) 


SIMULTANEOUS EQUATIONS. 175 


Exercise 63. 


Solve 
ea 4 Dirge 
= == 2 Sopa 
1 oe 7 re m 
BL vest x 
_OREAD Raveeyrts ch esting 
ie A Lg Oy 
es nm | 
blr ces eae Ca 
aoe s ans ‘ 
Fie wersintan Greaney 
Gy Ding 
Tica ie eS 
2 Gea Ol ac 
4 sty 3 | 10 ay 8 | 
hail a Se 
x mes ear ead) 
4 5 he 
5. —+—= 511 ll. ———=5 
este a ax by 
oi Ae eS od 
te 85y) 920 ax by J 
} 2 Oe dD 
6. —+—=8 12, —+~—= b 
Ce se) Fenny an 
set 3 


176 SCHOOL ALGEBRA. 


192. If three simultaneous equations are given, involv- 
ing three unknown numbers, one of the unknowns must be 
eliminated between two pazrs of the equations; then a 
second unknown between the two resulting equations. 

Likewise, if four or more equations are given, involving 
four or more unknown numbers, one of the unknowns must 
be eliminated between three or more pairs of the equations ; 
then a second between the pairs that can be formed of the 
resulting equations; and so on. 

Notg. The pairs chosen to eliminate from must be independent 


pairs, so that each of the given equations shall be used in the process 
of the eliminations. 


Solve: 24—8yt4z2= 4 (1) 
de+8y—Te= 12} (2) 
or— y—8z2= 5 ) 

Eliminate z between the equations (1) and (3). 


Multiply (1) by 2, 4x—6y+8z= 8 (4) 
(3) is 5e— y—8z= 5 
Add, 9x—Ty = 13 (5) 


Eliminate z between the equations (1) and (2). 

Multiply (1) by 7, 14a—21y 4+ 282 = 28 

Multiply (2) by 4, 122% +4 20y — 282 = 48 

Add, 262-~- y = 76 (6) 

We now have two equations (5) and (6) involving two unknowns, 
wand y. 


Multiply (6) by 7, 182% — Ty = 532 Pee, 
(5) is 9%—-Ty= 13 
Subtract (5) from (7), 173 a = 519 
*, Sed. 

Substitute the value of « in (6), 78 —y = 76. 

| “. ¥ = 2, 
Substitute the values of « and y in (1), 
: , 6—6442—4, 


"(g=1; 


SIMULTANEOUS EQUATIONS. 


Bi 


Exercise 64, 


. e+y— 8=0 
y+z—28=0 
z+z—14=0 


42-+3y+2z2= 25 
382 —2y+52= 20 
10%@— 5y+32=17 


5a— 2y—22=12 
z+ y+ z= 8 
12+ 8y+42= 42 


x—- yt z=ll 
82+ 38y—2z2=60 
1l0xz— 5y— 32> 
. l0oe¢— y+32=42 
Ta+t2y+ z2=51 
82+38y— z= 24 
. 9x +2y—3z=—160 
8a+9y+82=115 
224—s3y—52= 
. 62—2Qy+52=—53 
Sat 8yt+ Tz= 33 
et+ty+t z= 5 


62+ 2y—Tz= 5 


| 
| 


: Sah2y—te= 5 


22—- 


y + 82=405 


. 24+ T7y+10z2=25 
oe a ee IY Se 
T2—Ty—11lz=73 


| 


10. 


Ele 


in: 


13. 


14. 


15. 


16. 


ie 


18. 


 9a+18y—42= 


84a—b6y+ Tz=51 
4Av+t8y— 92= 538 


a+2y+10z= | 


Sey Te 


sz+3y+t Tz=3884 


2e+ y+ 2=256 


l0z= y+42+ 56) 
Co Mase mea 
242= x—dsy—18 


3x —dY— y— 2= 12} 


z=31 
x— yt v= 18| 


22+ 3y— 4z2=1 
10z— 6y+12z2=6 
e+l2y+ 22=5 
3sa+ by+22=3 
l2y+ 42—62= S| 
4) 


Dy —42—427==) 
s2+9y+ 2=9 
sa+t2y+ %2=204 
22— y+ 22= 264 | 
e+ y+10z=55 


2a+ yt2z=3 


SCHOOL ALGEBRA. 


178 


a re 
eS S =) 
| | | 

realS roils wayo 
| | 

SsIS Wala wWler 
2b SG ee 

eI AIS AIS 
ae) 

R 
S -) -) 
| | | 
sH rj =H 

rei 

a be + | 

AISA la ols 
| | ie 

HIS AIR AIar 
o 
ton | 


a Se 
or] — CO 
Yo) <H of 

lice el 
RIM ARLO alw 
+ + + 
ao SIH Sw 
+ + +4 
SIN Siod Sit 
< 

R 
a 
co oe) 2 
na) “A ox 

| | U 

alli ; 

ria ria + 
= | > 
ly O11 4S 
+ + + 


| 
es) =) > 
| | I 
CO] a N sH | CD 
| | | 
AIA N!arta 
eee ee 
N18 dla Wats 
19 
RX 
——____————,_ 
ae <H sH 
N cl 


I | | 


TOlr Olr la 


| + + 
AlL>HIR OID 


+ + | 


mi pea 13 fl & 


_ 
GR 


(ee eee 
ala 718 418 


! | | 


COLA Ol wtayr 


| | | 


HIS Nia tia 


R 
R 


CHAPTER X1L 


PROBLEMS INVOLVING TWO UNKNOWN 
NUMBERS. 


193, It is often necessary in the solution of problems to 
employ two or more letters to represent the numbers to be 
found. In all cases the conditions must be sufficient to 
give just as many equations as there are unknown numbers 
employed. 


194, If there are more equations than unknown numbers, 
some of them are superfluous or inconsistent; if there are 
less equations than unknown numbers, the problem is inde- 
terminate. 


(1) If A gives B $10, B will have three times as much 
money as A. If B gives A $10, A will have twice as much 
money as B. How much has each? 


Let x = number of dollars A has, 
and y = number of dollars B has. 


Then, after A gives B $10, 
x —10= the number of dollars A has, 
y + 10 = the number of dollars B has. 
“. yt 10=3(e—10). (1) 
If B gives A $10, 
x + 10 =the number of dollars A has, 


y — 10=the number of dollars B has. 
*.2+10=2(y—10). (2) 


From the solution of equations (1) and (2), # = 22, and y = 26. 
Therefore A has $22, and B has $26. 


180 SCHOOL ALGEBRA. 


(2) If the smaller of two numbers is divided by the 
greater, the quotient is 0.21, and the remainder 0.0057; 
but if the greater is divided by the smaller, the quotient 
is 4 and the remainder 0.742. Find the numbers. 


Let x = the greater number, 
and y = the smaller number. 
Then Ue ay (1) 
© 
and @— 0.742 _ 4. (2) 
- 
“. y — 0,21 x = 0.0057, (3) 
x—4y = 0.742. (4) 
Multiply (3) by 4, 4y — 0.84 2 = 0.0228 (5) 
(4) is —4y+ v= 0.742 
By adding, 0.16% = 0.7648 
“. = 4.78. 
Substituting the value of a in (4), 
—4y =— 4.038, 
“. y = 1.00985. 


Exercise 65. 


1. If A gives B $100, A will then have half as much 
money as B; but if B gives A $100, B will have one-third 
as much as A. How much has each? 


2. If the greater of two numbers is divided by the 
smaller, the quotient is 4 and the remainder 0.37; but if 
the smaller is divided by the greater, the quotient is 0.23 
and the remainder 0.0149. Find the numbers. 


3. A certain number of persons paid a bill. If there 
had been 10 persons more, each would have paid $2 less; 
but if there had been 5 persons less, each would have paid 
$2.50 more. Find the number of persons and the amount 


of the bill. 


PROBLEMS. 181 


4. A train proceeded a certain distance at a uniform 
rate. If the speed had been 6 miles an hour more, the time 
occupied would have been 5 hours less; but if the speed had 
been 6 miles an hour less, the time occupied would have 
been 74 hours more. Find the distance. 


Hint. If «=the number of hours the train travels, and y the 
number of miles per hour, then zy = the distance. 


5. A man bought 10 cows and 50 sheep for $750. He 
sold the cows at a profit of 10 per cent, and the sheep at a 
profit of 80 per cent, and received in all $875. Find the 


average cost of a cow and of a sheep. 


6. It is 40 miles from Dover to Portland. A sets out — 
from Dover, and B from Portland, at 7 o’clock A.M.,to meet — 
each other. A walks at the rate of 34 miles an hour, but 
stops 1 hour on the way; B walks at the rate of 24 miles an | 
hour. At what time of day and how far from Portland will 
they meet? 


7. The sum of two numbers is 85, and their difference 
exceeds one-fifth of the smaller number by 2. Find the 
numbers. 


8. If the greater of two numbers is divided by the 
smaller, the quotient is 7 and the remainder 4; but if three 
times the greater number is divided by twice the smaller, 
the quotient is 11 and the remainder 4. Find the numbers. 


9. If 3 yards of velvet and 12 yards of silk cost $60, 
and 4 yards of velvet and 5 yards of silk cost $58, what is 
the price of a yard of velvet and of a yard of silk? 


10. If 5 bushels of wheat, 4 of rye, and 3 of oats are sold 
for $9; 3 bushels of wheat, 5 of rye, and 6 of oats for $8.75; 
and 2 bushels of wheat, 3 of rye, and 9 of oats for $7.25; 
what is the price per bushel of each kind of grain? 


182 SCHOOL ALGEBRA. 


Nore I. A fraction the terms of which are unknown may be repre- 


sented by %. 
uy 


Ex. A certain fraction becomes equal to 4.if 2 is added 
to its numerator, and equal to 4 if 3 is added to its denomi- 
nator. Find the fraction. 








Let j = the required fraction. 
Then +2 =}, 
¥ 
and et Bast 1, 
y+s 


The solution of these equations gives 7 for x, and 18 for y. 


Therefore the required fraction is 7%. 


11. A certain fraction becomes equal to 4 if 3 is added 
to its numerator and 1 to its denominator, and equal to 4 
if 3 is subtracted from its numerator and from its denomi- 
nator. Find the fraction. 


12. A certain fraction becomes equal to 7% if 1 is addled 
to double its numerator, and equal to 4 if 3 is subtracted 
from its numerator and from its denominator. Find the 
fraction. 


13. Find two fractions with numerators 11 and 5 respec- 
tively, such that their sum is 14, and if their denominators 
are interchanged their sum is 24. 


14. There are two fractions with denominators 20 and 
16 respectively. The fraction formed by taking for a nu- 
merator the sum of the numerators, and for a denominator 
the sum of the denominators, of the given fractions, is equal 
to 4; and the fraction formed by taking for a numerator 
the difference of the numerators, and for a denominator the 
difference of the denominators, of the given fractions, 1s equal 
to $. Find the fractions, 


PROBLEMS. 183 


Nore If. A number consisting of two digits which are unknown 
may be represented by 10%+y, in which z and y represent the digits 
of the number. Likewise, a number consisting of three digits which 
are unknown may be represented by 1002 + 10y +z, in which g, y, 
and z represent the digits of the number. For example, the expres- 
sion 364 means 300 + 60 +4; or, 100 times 3 + 10 times 6 + 4. 


Ex. The sum of the two digits of a number is 10, and if 
18 is added to the number, the digits will be reversed. 
Find the number. 


Let x = tens’ digit, 
and *y = units’ digit. 
Then 10x + y = the number. 
Hence » x+y =10, (1) 
and 10a+y+18=10y +2. (2) 
From (2), 92—9y=—- 18, 
or e—y =— 2. (3) 
Add (1) and (3), Agia os 
and therefore a= 4, 
Subtract (3) from (1), 2y = 12, 
and therefore y= 6. 


Therefore the number is 46. 


15. The sum of the two digits of a number is 9, and if 27 
is subtracted from the number, the digits will be reversed. 
Find the number. 


16. The sum of the two digits of a number is 9, and if 
the number is divided by the sum of the digits, the quotient 
is 5. Find the number. 


17. A certain number is expressed by two digits. The 
sum of the digits is 11. If the digits are reversed, the new 
number exceeds the given number by 27. Find the number. 


18. A certain number is expressed by three digits. The 
sum of the digits is 21. The sum of the first and last digits 
is twice the middle digit. Ifthe hundreds’ and tens’ digits 
are interchanged, the number is diminished by 90. Find 
tuc number. 


184 SCHOOL ALGEBRA. 


19. A certain number is expressed by three digits, the 
units’ digit being zero. If the hundreds’ and tens’ digits are 
interchanged, the number is diminished by 180. If the 
hundreds’ digit is halved, and the tens’ and units’ digits are 
interchanged, the number is diminished by 336. Find the 
number. 


20. A number is expressed by three digits. If the digits 
are reversed, the new number exceeds the given number by 
99. Ifthe number is divided by nine times the sum of its 
digits, the quotient is 8. The sum of the hundreds’ and 
units’ digits exceeds the tens’ digit by 1. Find the number. 


Nore III. If a boat moves at the rate of w miles an hour in still 
water, and if it is on a stream that runs at the rate of y miles an 


hour, then, + y represents its rate down the stream, 


x — y represents its rate up the stream. 


21. A boatman rows 20 miles down a river and back in 
8 hours. He finds that he can row 5 miles down the river 
in the same time that he rows 3 miles up the river. Find 
the time he was rowing down and up respectively. 


22. A boat’s crew which can pull down a river at the 
rate of 10 miles an hour finds that it takes twice as long to 
row a mile up the river as to row a mile down. Find the 
rate of their rowing in still water and the rate of the 
stream. 


23. A boatman rows down a stream, which runs at the 
rate of 24 miles an hour, for a certain distance in 1 hour 
and 80 minutes; it takes him 4 hours and 80 minutes to 
return. Find the distance he pulled down the stream and 
his rate of rowing in still water. 


Note IV. It is to be remembered that if a certain work can be 
done in « units of time (days, hours, etc.), the part of the work done 


in one unit of time will be represented by B 
x 


PROBLEMS. 185 


Ex. A cistern has three pipes, A, B, and OC. A and B 
will fill the cistern in 1 hour and 10 minutes, A and C in 
1 hour and 24 minutes, B and C in 2 hours and 20 minutes. 
How long will it take each pipe alone to fill it? 

1 hour and 10 minutes = 70 minutes, 


1 hour and 24 minutes= 84 minutes, 
2 hours and 20 minutes = 140 minutes. 


Let x = number of minutes it takes A to fill it, 
y = number of minutes it takes B to fill it, 
and ’ g=number of minutes it takes C to fill it. 
Then 1 1 dye the parts A, B, and C can fill in one minute 
oie ta respectively, 
and : + ; = the part A and B together can fill in one minute. 
But a =the part A and B together can fill in one minute. 
tein () 
sie honey 8) 
In like manner, L + Lois (2) 
LM ae Pose 
Wy SN 
d as Ca Yecy nd Ee 3 
an a tag (3) 
Add, and divide by 2, 14141-4. (4) 
a agi OU) 
Tuy) 1 
Subtract (1) from (4), er ith 
ubtract (1) from (4) arr 
Subtract (2) from (4), ae 
y 210 
Subtract (3) from (4), Teas 
pe 9) 


Therefore a, y, 2 = 105, 210, 420, respectively. 
Hence A can fill it in 1 hour and 45 minutes, B in 3 
hours and 380 minutes, and C in 7 hours. 


24. A and B can doa piece of work together in 3 days, 
A and C in 4 days, B and C in 44 days. How long will it 
take each alone to do the work? 


186 SCHOOL ALGEBRA. 


25. A and B can doa piece of work in 24 days, A and 
C in 34 days, B and C in4 days. How long will it take 
each alone to do the work? 


26. A and B can do a piece of work in a days, A and C 
in 6 days, B and C ine days. How long will it take each 
alone to do the work ? 


Notre V. If represents the number of linear units in the length. 
and y in the width, of a rectangle, wy will represent the number ot! 
its units of surface; the surface unit having the same name as the 
linear unit of its side. 


27. If the length of a rectangular field were increased 
by 5 yards and its breadth increased by 10 yards, its area 
would be increased by 450 square yards; but if its length 
were increased by 5 yards and its breadth diminished by 
10 yards, its area would be diminished by 350 square yards. 


Find its dimensions. 


28. If the floor of a certain hall had been 2 feet longer 
and 4 feet wider, it would have contained 528 square feet 
more; but if the length and width were each 2 feet less, it 
would contain 316 square feet less. Find its dimensions. 


29. If the length of a rectangle was 4 feet less and the . 
width 3 feet more, the figure would be a square of the same 
area as the given rectangle. Find the dimensions of the 
rectangle. 


Norge VI: In considering the rate of increase or decrease in quan- 
tities, it is usual to take 100 as a common standard of reference, so 
that the increase or decrease is calculated for every 100, and there- 
fore called per cent. 

It 1s to be observed that the representative of the number result- 
ing after an increase has taken place is 100 + increase per cent; and 
after a decrease, 100 — decrease per cent. 

Interest depends upon the time for which the money is lent, as 


PROBLEMS. 187 


well as upon the rate per cent charged, the rate per cent charged 
being the rate per cent on the principal for one year. Hence, 


Simple interest = Principal x sated er cent x Time 


where Time means number of years or fraction of a year. 


Amount = Principal + Interest. 


In questions relating to stocks, 100 is taken as the representative 
ot the stock, the price represents its market value, and the per cent 
represents the interest which the stock bears. Thus, if six per cent 
stocks are quoted at 108, the meaning is, that the price of $100 of 
the stock is $108, and that the interest derived from $100 of the 
stock will be 78, of $100, that is, $6 a year. The rate of interest on 
the money invested will be 42° of € per cent. 


30. A man has $10,000 invested. For a part of this 
sum he receives 5 per cent interest, and for the rest 6 per 
cent; the income from his 5 per cent investment is $60 
more than from his 6 per cent. How much has he in each 
investment ? 


31. A sum of money, at simple interest, amounted in 4 
years to $ 29,000, and in 5 years to $30,000. Find the sum 
and the rate of interest. 


32. A sum of money, at simple interest, amounted in 10 
months to $2100, and in 18 months to $2180. Find the 
sum and the rate of interest. 


33. A person has a certain capital invested at a certain 
rate per cent. Another person has $2000 more capital, 
and his capital invested at one per cent better than the 
first, and he receives an income of $150 greater. A third 
person has $8000 more capital, and his capital invested at 
two per cent better than the first, and he receives an income 
of $280 greater. Find the capital of each and the rate at 
which it is invested. 


188 SCHOOL ALGEBRA. 


34. A sum of money, at simple interest, amounted in m™ 
years to ¢ dollars, and in 7 years to d dollars. Find the 
sum and the rate of interest. 


35. A sum of money, at simple interest, amounted in m 
months to @ dollars, and in m months to 0 dollars. Find 
the sum and the rate of interest. 


36. A person has $18,375 to invest. He can buy 3 per 
cent bonds at 75, and 5 per cent bonds at 120. How much 
of his money must he invest in each kind of bonds in order 
to have the same income from each investment? 


Hint. Notice that the 3 per cent bonds at 75 pay 4 per cent on 
the money invested, and 5 per cent bonds at 120 pay 43 per cent. 


37. A man makes an investment at 4 per cent, and a 
second investment at 44 per cent. His income from the 
two investments is $715. If the first investment had been 
at 44 per cent and the second at 4 per cent, his income would 
have been $730. Find the amount of each investment. 


(1) In a mile race A gives B a start of 20 yards and 
beats him by 80 seconds. At the second trial A gives Ba 
start of 32 seconds and beats him by 935 yards. Find the 
number of yards each runs a second. 

Let 2 = number of yards A runs a second, 
and y = number of yards B runs a second. 


Since there are 1760 yards in a mile, 


ve = number of seconds it takes A to run a mile. 


Since B has a start of 20 yards, he runs 1740 yards the first trial; 
and as he was 30 seconds longer than A, 


a + 30 =the number of seconds B was running. 


1780 = the number of seconds B was running. 


/ 170 1760 a) 
y x 


PROBLEMS. 189 


In the second trial B runs 1760 — 935, = 1750;%, yards. 
_ 175038 1760 
EN tae aE 


y 
From the solution of equations (1) and (2), r = 518, and y = 544. 


32. (2) 


Therefore A runs 543 yards a second, and B runs 5,3, 
yards a second. 


(2) A train, after travelling an hour from A towards B, 
meets with an accident which detains it half an hour; after 
which it proceeds at four-fifths of its usual rate, and arrives 
an hour and a quarter late. If the accident had happened 
30 miles farther on, the train would have been only an hour 
late. Find the usual rate of the train. 


Since the train was detained 4 an hour and arrived 11 hours late, 
the running time was 3 of an hour more than usual. 


Let y = number of miles from A to B, 
and 5a = number of miles the train travels per hour. 
Then y—5a#=number of miles the train has to go after the 
accident. 
Hence ae = number of hours required usually, 
© 
and woes = number of hours actually required. 
© 
UE OR SY 9 loan it hours of running time. 
4a 5x 
But 2 == loss in hours of running time. 
. y¥—se y-de_ 3 (1) 
Ag 5a 4 


If the accident had happened 30 miles farther on, the remainder 
of the journey would have been y—(5a + 30), and the loss in running 
time would have been } an hour. 


r y—(6e2+30) y—(Sr+30) 1 (2) 
; 4a 5a Z 


From the solution of equations (1) and (2), «= 6, and 5a = 30¢ 


Therefore the usual rate of the train is 80 miles an hour. 


190 SCHOOL ALGEBRA. 


38. Two men, A and B, run a mile, and A wins by 2 
seconds. In the second trial B has a start of 18} yards, 
and wins by 1 second. Find the number of yards each 
runs a second, and the number of miles each would run in 
an hour. 


39. In a mile race A gives B a start of 3 seconds, and is 
beaten by 124 yards. In the second trial A gives B a start 
of 10 yards, and the race is a tie. Find the number of 
yards each runs a second. At this rate, how many miles 
could each run in an hour? 


40. In amile race A gives B a start of 44 yards, and is 
beaten by 1 second. Ina second trial A gives B a start of 
6 seconds, and beats him by 9% yards. Find the number 
of yards each runs a second. 


41. An express train, after travelling an hour from A 
towards B, meets with an accident which delays it 15 min- 
utes. It afterwards proceeds at two-thirds its usual rate, 
and arrives 24 minutes late. If the accident had happened 
5 miles farther on, the train would have been only 21 
minutes late. Find the usual rate of the train. 


42. A train, after running 2 hours from A towards B, 
meets with an accident which delays it 20 minutes. It 
afterwards proceeds at four-fifths its usual rate, and arrives 
1 hour and 40 minutes late. If the accident had happened 
40 miles nearer A, the train would have been 2 hours late. 
Find the usual rate of the train. 


43. A and B can doa piece of work in 23 days, A and 
C in 84 days, B and C in 32 days. In what time can all 
three together, and each one separately, do the work? 


44. A sum of money, at interest, amounts in 8 months 
to $1488, and in 15 months to $1580. Find the principal 


>and the rate of interest, 


PROBLEMS. 19] 


45. A number is expressed by two digits, the units’ digit 
being the larger. If the number is divided by the sum of 
_its digits, the quotient is 4. If the digits are reversed and 

the resulting number is divided by 2 more than the differ- 
~ ence of the digits, the quotient is 14. Find the number. 


46. A and B together can dig a well in 10 days. They 
work 4 days, and B finishes the work in 16 days. How 
long would it take each alone to dig the well? 


47. The denominator of the greater of two fractions is 
20. The fraction formed by taking for a numerator the 
sum of the numerators of the two fractions, and for a 
denominator the sum of the denominators, is equal to 2. 
The fraction similarly formed with the difference of the 
numerators, and of the denominators, is equal to 4. The 
sum of the numerators is twice the difference of the denomi- 
nators. Find the fractions. 


48. A cistern can be filled in 5 hours by two pipes, A 
and B, together. Both are left open for 3 hours and 45 
minutes, and then A is shut, and B takes 3 hours and 45 
minutes longer to fill the cistern. How long would it take 
each pipe alone to fill the cistern ? 


49. A man put at interest $20,000 in three sums, the 
first at 5 per cent, the second at 44 per cent, and the 
third at 4 per cent, receiving an income of $905 a year. 
The sum at 44 per cent is one-third as much as the other 
two sums together. Find the three sums. 


50. An income of $385 a year is obtained from two in- 
vestments, one in 44 per cent stock and the other in 5 per 
cent stock. If the 44 per cent stock should be sold at 
110, and the 5 per cent at 125, the sum realized from both 
stocks together would be $8300. How much of each stock 
is there? 


192 SCHOOL ALGEBRA. 


51. A boy bought some apples at 3 for 5 cents, and 
some at 4 for 5 cents, paying for the whole $1. He sold 
them at 2 cents apiece, and cleared 40 cents. How many 


of each kind did he buy ? 


52. Find the area of a rectangular floor, such that if 3 
feet were taken from the length and 3 feet added to the 
breadth, its area would be increased by 6 square feet, but 
if 5 feet were taken from the breadth and 38 feet added to 
the length, its area would be diminished by 90 square feet. 


53. A courier was sent from A to B, a distance of 147 
miles. After 7 hours, a second courier was sent from A, 
- who overtook the first just as he was entering B. The time 
required by the first to travel 17 miles added to the time 
required by the second to travel 76 miles is 9 hours and 
40 minutes. How many miles did each travel per hour? 


-54. A box contains a mixture of 6 quarts of oats and 9 
of corn, and another box contains a mixture of 6 quarts of 
oats and 2 of corn. How many quarts must be taken from — 
each box in order to have a mixture of 7 quarts, half oats 
and half corn? 


55. A train travelling 30 miles an hour takes 21 minutes 
longer to go from A to B than a train which travels 36 
miles an hour. Find the distance from A to B. 


56. A man buys 570 oranges, some at 16 for 25 cents, 
and the rest at 18 for 25 cents. He sells them all at the 
rate of 15 for 25 cents, and gains 75 cents. Hew many of 
each kind does he buy ? 


57. A and B run a mile race. In the first heat B 
receives 12 seconds start, and is beaten-by 44 yards. In 
the second heat B receives 165 yards start, and arrives at 
the winning post 10 seconds before A. Find the time in 
which each can run a mile. 


PROBLEMS. 198 


hi MP AKO i 
“ 


INDETERMINATE PROBLEMS. 


195. If a single equation is given which containstwo 
unknown numbers, and no other condition is imposed, the 
number of its solutions is wnlimited; for, if any value be 
assigned to one of the unknowns, a corresponding value 
may be found for the other. Such an equation is said to 
be indeterminate. 


196. The values of the unknown numbers in an inde- 
terminate equation are dependent upon each other ; so that, 
though they are unlimited in number, they are confined to 
a particular range. 

This range may be still further iene by requiring these 
values to satisfy some given condition; as, for instance, that 
they shall be positive integers. With such restrictions the 
equation may admit of a definite number of solutions. 


Ex. A number is expressed by two digits. If the num- 
ber is divided by the sum of its digits diminished by 4, the 
quotient is 6. Find the number. 


The single statement is 


ME A eae 

e+y—4 
Whence 4a =5y — 24, 
and w=y+%—6 
atte 
=y 645 


We see from + that the values of y which will be integral are 4, 8, 


y 
4 
that the least positive integral value of y which will give to x a posi- 


tive integral value, is 8. If we put 8 for y in (1), we find <=4. Hence 
the nunber required is 48. 


12, 16, or some Rie multiple of 4, and from the relation z=y—6 +7 


194. SCHOOL ALGEBRA. 


Exercise 66. 


1. A number is expressed by two digits. Ifthe number 
is divided by the last digit, the quotient is 15. Find the 


number. 


2. A number is expressed by three digits. The sum of 
the digits is 20. If 16 is subtracted from the number and 
the remainder divided by 2, the digits will be reversed. 
Find the number. 


Here e+y+z2= 20, 
and sete Lah BE 
Eliminate y and reduce, and we have 
42 =72 + 8. 


3. A man spends $114 in buying calves at $5 apiece, 
and pigs at $3 apiece. How many did he buy of each? 


4. In how many ways can a man pay a debt of $87 
with five-dollar bills and two-dollar bills? 


5. Find the smallest number which when divided by 5 
or by 7 gives 4 for a remainder. 


n—4 nm—4 


Let n =the number, then =x, and arr a Up 





6. A farmer sells 15 calves, 14 lambs, and 18 pigs for 
$200. Some days after, at the same price, he sells 7 calves, 
11 lambs, and 16 pigs, for which he receives $141. What 
was the price of each? 


Discussion OF PROBLEMS. 


197. The discussion of a problem consists in making 
various suppositions as to the relative values of the given 
numbers, and explaining the results. We will illustrate by 
an example: 


PROBLEMS. 195 


Two couriers were travelling along the same road, and in 
the same direction, from C towards D. A travels at the 
rate of m miles an hour, and B at the rate of » miles an 
hour. At 12 o’clock B was d miles in advance of A. When 
will the couriers be together ? 


Suppose they will be together x hours after 12. Then A has trav- 
elled mz miles, and B has travelled nz miles, and as A has travelled 
d miles more than B 

me =nx +d, 
or mz — nx = d. 


d 


m—n 


._ t= 





Discussion oF THE Prosiem. 1. If m is greater than n, the value 





of a, namely, d , 18 positive, and it is evident that A will over- 
m—n 


take B after 12 o'clock. 


2. If m is less than n, then 


B travels faster than A, and as he is d miles ahead of A at 12 o'clock, 
it is evident that A cannot overtake B after 12 o'clock, but that B 


passed A before 12 o’clock by g 
n— 


fore, that the couriers were together after 12 o'clock was incorrect, 
and the negative value of # points to an error in the supposition. 


will. be negative. In this case 





hours. The supposition, there- 








3. If m equals n, then the value of a, that is, , assumes the 


m—n 
form < Now if the couriers were d miles apart at 12 o’clock, and if 
they had been travelling at the same rates, and continue to travel at 
the same rates, it is obvious that they never had been together, and 


that they never will be together, so that the SEs = may be regarded 
as the symbol of impossibility. 





4. If m equals n and d is 0, then becomes >. Now if the 


—n 
couriers were together at 12 o'clock, and if they had been travelling 
at the same rates, and continue to travel at the same rates, it is 
obvious that they had been together all the time, and that rach will 


continue to be together all the time, so that the symbel 2 @ may be 
regarded as the symbol of indetermination. 


196 SCHOOL ALGEBRA. 


Exercise 67. 


1. A train travelling 6 miles per hour is m hours in 
advance of a second train which travels a miles per hour. 
In how many hours will the second train overtake the first ? 

bm 
Ans. 
a—b 
Discuss the result (1) when a>b; (2) when a=); (3) whena<b. 





2. A man setting out on a journey drove at the rate of 
a miles an hour to the nearest railway station, distant d 
miles from his house. On arriving at the station he found 
that the train for his place of destination left ¢ hours before. 
At what rate should he have driven in order to reach the 
station just in time for the train? i, 


Ans. 





b— ae 
Discuss the result (1) when e=0; (2) when one, (3) when 
en 2. In case (2), how many hours did the man have to drive 


from his house to the station? In case (3), what is the meaning of 
the negative value of ¢? 


3. A wine merchant has two kinds of wine which he sells, 
one at a dollars, and the other at 6 dollars per gallon. He 
wishes to make a mixture of J gallons, which shall cost him 
on the average m dollars a gallon. How many gallons 
must he take of each? 


Pin laa - of the first ; ie) = er nego 
—_— a = 

Discuss the question (1) when a=6; (2) when a or b=™m; (3) 
when a=b=m; (4) when a>b and <™m; (5) when a>b and 
b>m. 


CHAPTER XIII. 
INEQUALITIES. 


198. An inequality consists of two unequal numbers 
connected by the sign of inequality. Thus, 12>4 and 
4 < 12 are inequalities. 


199. Two inequalities are said to be of the same direction 
if the first members are both greater or both less than the 
second members; that is, if the signs of inequality point in 
the same direction. 


200. Two inequalities are said to be the reverse of each 
other if the signs point in opposite directions. 


201, If equal numbers are added to, or subtracted from, 
the members of an inequality, the inequality remains in 
the same direction. Thus, ifa>4, thena+c>6-+e, and 
a—e>b—e. Hence, 

A term can be transposed from one member of an in- 
equality to the other without altering the inequality, provided 
us sign ws changed. 


202. If unequals are taken from equals, the result is an 
inequality which is the reverse of the given inequality. 
Thus, if~=y, anda> 6), then x—-a<y—b. 


208, If the signs of the terms of an inequality are 
changed, the inequality is reversed. Thus, if a> 6, then 
—a<—b. (See § 33.) 


CHAPTER XIV. 


INVOLUTION AND EVOLUTION. 


205. The operation of raising an expression to any re- 
quired power is called Involution, 

Every case of involution is merely an example of multi- 
plication, in which the factors are equal. 


206. Index Law. If mis a positive integer, by definition 


Oa aK Ge: to m factors. 


Consequently, if m and 7 are both positive integers, 
(a")" = a" X a" X a” -:--- to m factors 
=(aXa---- ton factors)(a X a «++: to n factors) 
DAG taken m times 
= GK OR ae to mn factors. 
eat hier 


This is the index law for involution. 


Eeirmed 80, (try) =e oe 

And (ab)"= ab x ab + to n factors 
= (aX a+ ton factors)(d x b+. to m factors) 
=a 0-0" 


208. If the exponent of the required power is a composite 
number, the exponent may be resolved into prime factors, 
the power denoted by one of these factors found, and the 
result raised to a power denoted by a second factor of the 
exponent; and soon. ‘Thus, the fourth power may be ob- 
tained by taking the second power of the second power ; 


INVOLUTION AND EVOLUTION. 201 


the sixth by taking the second power of the third power ; 
and so on. 


209. From the Law of Signs in multiplication it is evi- 
dent that all even powers of a number are positiwe; all odd 
powers of a number have the same sign as the number itself. 

Hence, no even power of any number can be negative ; 
and the even powers of two compound expressions which 
have the same terms with opposite signs are identical. 


Thus, (6—a)?’=§{—(a—d)??=(a-— }). 


210. Binomials. By actual multiplication we obtain, 
(a+ b’=a@+2ab+0'; 
(a+ 6 = a+ 3076 + 8ab?+ 8B; 
(a -+ 6)*== a+ 40°) + 6070? + 408? + OF. 

In ihese results it will be observed that: : 


I. The number of terms is greater by one than the ex- 
ponent of the power to which the binomial is raised. 

II. In the first term, the exponent of a is the same as 
the exponent of the power to which the binomial is raised ; 
and it decreases by one in each succeeding term. 

III. 4 appears in the second term with 1 for an exponent, 
and its exponent increases by 1 in each succeeding term. 

IV. The coefficient of the first term is 1. 

V. The coefficient of the second term is the same as the 
exponent of the power to which the binomial is raised. 

VI. The coefficient of each succeeding term is found 
from the next preceding term by multiplying the coefficient 
of that term by the exponent of a, and dividing the product 
by a number. greater by one than the exponent of 8. 


If } is negative, the terms in which the odd powers of 6 
occur are negative. Thus, 


202 - SCHOOL ALGEBRA. 


(1) (a — 6)? = a’ — 807) 4 38ab?— b. 

(2) (a — b)*= at — 40°) + 6070’ — 4ab*+ Bt. 

By the above rules any power of a binomial of the form 
a-+- 6 may be written at once. 

Nore. The double sign + is read plus or minus; and a+b means 
dat) OG — 0, 

211. The same method may be employed when the terms 
of a binomial have coefficients or exponents. 

Since (a— b?=ai— 3a’) + 3ab’— Bb’, 
putting 52” for a, and 27° for 6, we have 

(52° —2y)), , 
= (52%) — 8 (52°)(2y") + 8(62°)(2y)'— (2y), 
= 1252°— 150 .x'y* + 60 2*y* — 87’. 
Since (a@—b)*=at— 4a°b 4 6a°b*?— 4ab* + BF, 
putting 2? for a, and dy for 6, we have ~ 
(0 — 4), 
= (2°) 4(2*)"hy) + 6(0°)" hy) 40°(hy)+ by) 
= a — 2aby + paly’ — Fey + Tey’. 

212. In like manner, a polynomial of three or more terms 
may be raised to any power by enclosing its terms in paren- 
theses, so as to give the expression the form of a binomial. 
Thus, 

Q) @4+o+eP=[at +o}, 
7 =a+3e(6+c)+3a(6+cP?+(6+c), 
=a+3@b+3we+8ab’+ babe 
+ 38a?+0?+30?e+86e' +e, 


ro 


Ler 


MD chin ah) etre ty 


INVOLUTION AND EVOLUTION. 203 


(2) (@—22°+ 32+ 4)’, 

= [(e? — 20°) + (82 +49, 
= (2° — 22")? + 2 (a? — 227) (824 4)+ (824+4), 

= 2 —49°+ 4244+ 62'—423—16 474 9274 242+16, 
= #— 42° 4+ 102*—42°— 7274+ 2424-16. 


Exercise 69. 


Raise to the required power : 


iis 


2. 


3. 


4. 


(a')’. 
C292 
3.ab° 
(—5ab*c*)*. 
(— T2’y2*). 


3 abi ct 5 
& oar 


(— 2zy'*)*. 
(— 80°b*z*)’. 





327y° 4 
(ae) 


10. 


(2 + 2). 


11. (a? — 2), 

12. («+3)*. 

13. (22+1)°. 

14. (2m’— 1)’. 

15. (2a+3y). 

16. (2x—y)*. 

17. (ay — 2)". 

Lhe Clee ay 
19. (1— 224327) 
20. (l—a-+a’)’. 


EVOLUTION. 


. (84a 4+ 5%), 


218. The mth root of a number is one of the n equal 


factors of that number. 


The operation of finding any required root of an expres- 


sion is called Evolution. 


204 _ SCHOOL ALGEBRA. 


Every case of evolution is merely an example of factor- 
ang, in which the required factors are all equa/. Thus, the 
square, cube, fourth, ..... roots of an expression are found 
by taking one of its two, three, four ..... equal factors. 

The symbol which denotes that a square root is to be 
extracted is »/; and for other roots the same symbol is 
used, but with a figure written above to indicate the root; 
thus, ¥/, ~/, etc., signifies the third root, fourth root, ete. 


214, Index Law. If m and 7 are positive integers, we 
have, (§ 206), 
(ey ane. 
Consequently, oe eu 
Thus, the cube root of a® is a’; the fourth root of 8la” 
is 8a*; and so on. 
This is the index law for evolution. 


215, Also, since (ad)"= ad”, 
conversely, nO Cle Ne ay 
and Vb aN ON 


Hence, to find the root of a simple expression : 


Divide the exponent of each factor by the dex of the root, 
and take the product of the resulting factors. 


216. From the Law of Signs it is evident that 


I. Any even root of a positive number will have the 
double sign, +. 

II. There can be no even root of a negative number. 

For V—2’ is neither +a nor —2; since the square of 
+2=-+2’, and the square of —r=+ 2’. 

The indicated even root of a negative number is called 
an imaginary number. 

III. Any odd root of a number will have the same sign 
as the number. 


INVOLUTION AND EVOLUTION. 205 


Thus, \ieaa ees: V— 27min? = — 3 mn’; 
187’ 9y 


V 





16.2%y” or 2 wy 
8lai® ita ore 


217. If the root of a number expressed in figures is not 
readily detected, it may be found by resolving the number 
into its prime factors. Thus, to find the square root of 
3,415,104: 

23| 3415104 
2?} 426888 
rok 





1l 


3,415,104 = 2°x 3? x 7? x 112, 
~, V3,415,104 = 22x 83 x7 x 11 = 1848. 


Exercise 70. 





Simplify : 
eV 4277/: Pee 160. 17 | dao 
‘ ry 
2, 642°. 10. V7292%. ee 





; 3, ,6 
3. V16a%y” 11. W243 752% 18. Ne 
: a ae tows 27 2° 


4. W— 32a”. 12. V—1728d'. a 5 32a. 
EDT. ay = 84apt. pee 





a 4737 0 Awe 
6. V 25a". 14. V8la*. 20. Sean 
EET TE: 15. VW512a"b", seese 
Se 21. ae 
8. V64z". 16. -Vaimy?m, 216 a 





206 SCHOOL ALGEBRA. 


SQuaRE Roots oF CoMPOUND EXPRESSIONS. 


218. Since the square of a+6 is a’+2ab+0’, the 
square root of a?+ 2ab+ 0 is a+. 

It is sa to find a method of extracting the root 
a+6 when a’+ 2ab + @ is given: 


Ex. The first term, a, of the root is obviously the square root of 
the first term, a”, in the expression. 
a+2ab+bla+b If the a? be subtracted from the given 


a? expression, the remainder is 2ab + b?, 
2a+6| 2ab+ 0? Therefore the second term, b, of the root 
2004 7 is obtained when the first term of this | 


remainder is divided by 2a, that is, by 
double the part of the root already found. Also, since 


2Qab + b2 =(2a+ b)d, 


the divisor is completed by adding to the trial-divisor the new term of 
the root. 


1) Find the square root of 252?— 20a°y + 42'y’. 
q y Y 


25 x? — 20 a8y + 4aty?|5a — 2a%y 
25 a” 


10% — 22°y|— 20a%y + 4aty? 
— 20 a%y + 4aty? 


The expression is arranged according to the ascending powers of z. 

The square root of the first term is 5a, and 52 is placed at the 
right of the given expression, for the first term of the root. 

The second term of the root, —2a%y, is obtained by dividing 
— 20a3y by 102, and this new term of the root is also annexed to the 
divisor, 102, to complete the divisor. 


219. The same method will apply to longer expressions, 
if care be taken to obtain the ¢ria/-divisor at each stage of 
the process, by doubling the part of the root already found, 
and to obtain the complete divisor by annexing the new term 
of the root to the trial-divisor. 


INVOLUTION AND EVOLUTION. 207 


Ex. Find the square root of 
1+- 1027+ 252*+ 162° — 2445 — 202° — 42. 


16 2° ~— 2405 + 252 — 2023 + 1027-42 + 1[423— 3227+ 2e—1 
16 26 





82? — 3.a?|— 2405 + 25 ot 
— Mo + 924 


82? — 62? + 2x] 16 at — 2023 + 102? 
162a¢— 12434 42? 


8a — 6074+ 4e2—1]/— 827+ 62?—4241 
— 823+ 627-4241 


The expression is arranged according to the descending powers of z. 
It will be noticed that each successive trial-divisor may be obtained 
by taking the preceding complete divisor with its last term doubled. 


Exercise 71. 


* 


Find the square root of 
ib Cady jyieet ya Le 
2. 9at— 6a? + 18a?—4a+-4. 
3. 4a* —12a%y + 29277? — 80 ay? + 2574. 
4. 1442+ 102? + 122' + 92". 
5. 16—962+ 2162? — 2162°+ 81 2+. 
6. z*—222°4+ 952?+ 2862+ 169. 
7. 42*—1ll2’?+ 25 — 122'°-+ 30-2. 
8 9a +49 — 122° — 282 + 462° 
9. 49a*+ 12627 + 121 — 732? — 1982. 
10. 162*—30x2— 81a’ + 242° + 25. 


208 SCHOOL ALGEBRA. 


1 
12. 42*+ 42% —1xe+ 16 
cae 


a? 


4a? 
Basar ae 


OO oe 1 
Vite smowcein tii Ue 
fee Co laa sa ant 


Ao? ot 4h eo : 
16,0 — : 
tie eaaie ogee 


2 
17. fan 28 4 at Ba +2 


18. 16a*+189°y+8a'+4y7+4y+1. 


Gat _ 92", 482" EF aia 
4 2 4 2 4 








19. 


Tie yan ot aah ESE E,? 
a a 


Find to three terms the square root of 


21. a+b. 24. lta. 27. 427+ 3. 
22. atly. 25. 1— 2a. 28. 4—3a. 
23. 1+ 2a. 26. 4a°+ 20. 29. 4a?—1. 


220. Arithmetical Square Roots. In the general method 
of extracting the square root of a number expressed by 
figures, the first step is to mark off the figures in groups. 

Since 1 = 1?, 100=10?, 10,000 = 1007, and so on, it is 
evident that the square root of any number between 1 and 
100 lies between 1 and 10; the square root of any number 
between 100 and 10,000 lies between 10 and 100. In 


INVOLUTION AND EVOLUTION. 209 


other words, the square root of any number expressed by 
one or two figures is a number of one figure; the square root 
of any number expressed by ¢hree or four figures is a num- 
ber of éwo“figures; and so on. 

If, therefore, an integral square number is divided into 
groups of two figures each, from the right to the left, the 
number of figures in the root will be equal to the number 
of groups of figures. The last group to the left may have 
only one figure. 


Ex. Find the square root of 3249. 


32 49 (57 In this case, a in the typical form a? + 2ab + 0? 
20 <= represents 5 tens, that is, 50, and 5b represents 7. 

107) 749 The 25 subtracted is really 2500, that is, a?, and the 
Tee complete divisor 2a +b is 2x 50+ 7 = 107. 


221, The same method will apply to numbers of more 
than two groups of figures by considering a in the typical 
form to represent at each step the part of the root already 
found. 

It must be observed that a represents so many tens with 
respect to the next figure of the root. 


Ex. Find the square root of 5,322,249. 


5 32 22 49 (2307 
- hc 
43) 132 
129 
4607) 32249 
32249 


222. If the square root of a number has decimal places, 
the number itself will have twice as many. ‘Thus, if 0.21 is 
the square root of some number, this number will be (0.21)? 
= 0.21 x 0.21 = 0.0441; and if 0.111 be the root, the num- . 
ber will be (0.111)? = 0.111 x 0.111 = 0.012321. 


210 SCHOOL ALGEBRA. 


Therefore, the number of decimal places in every square 
decimal will be even, and the number of decimal places in 
the root will be Aa/f as many as in the given number itself. 

Hence, if a given number contain a decimal, we divide 
it into groups of two figures each, by beginning at the 
decimal point and marking toward the left for the integral 
number, and toward the right for the decimal. We must 
be careful to have the last group on the right of the deci- 
mal point contain two figures, annexing a cipher when 





necessary. 
Ex. Find the square roots of 41.2164 and 965.9664. 
41.21 64 (6.42 9 65.96 64 (31.08 

36 Do 

124) 521 61) 65 

496 61 
1282) 2564 6208) 49664 
2564 49664 





223. If a number contain an odd number of decimal 
places, or if any number give a remainder when as many 
figures in the root have been obtained as the given number 
has groups, then its exact square root cannot be found. We 
may, however, approximate to its exact root as near as we 
please by annexing ciphers and continuing the operation. 

The square root of a common fraction whose denominator 
is not a perfect square can be found approximately by 
reducing the fraction to a decimal and then extracting the 
root; or by reducing the fraction to an equivalent fraction 
whose denominator is a perfect square, and extracting the 
square root of both terms of the fraction. Thus, 


pee cote 
NE = \/0.625 = 0.79057 ; 


or 4[2= Se 


Mira ey yet ae 


INVOLUTION AND EVOLUTION. 


211 


Ex. Find the square roots of 3 and 357.357. 


ede ev sate 
1 

27) 200 
189 

343) 1100 
1029 


3462) 7100 
6924 





3.57.35 70 (18.903..... 


1 


28) 257 
244 
369) 3335 

3321 





37803) 147000 


Exercise 72. 


Find the square root of 


meee. 


1 
2 
3. 12544. 
4. 253009. 
5 


~ 529984. 10. 


6. 
. 1225. es 
8 
9 


150.0625. 
118.1569. 


. 172.3969. 
- 9200.140544. 
1303.282201. 


113409 





11. 
12. 
iS: 
14. 


HAS 


640.343025, 
100.240144. 
316.021729. 
454.585041. 
-§127.276025. 


Find the square root to four decimal places of: 


16. 10. Cpe tay 
17 wo. 20. 0.7. 


18. 5. SMS ES 


OF sal).00 Fe 
Peas, Oe eo 
24-0681" 


25. 
26. 


27. 


224, Cube Roots of Compound Expressions. 
of a+6 is a®+ 30°) + 3ab?+ 6’, the cube root of 


2, 08. fo, 
UL a 
930; ad 


Since the cube 


a+ 8a7b + 38ab?+ & is a+. 


It is required to devise a method for extracting the cube 
root a+ 6 when a*+ 3a7b + 3a6’+ 6° is given: 


a2 SCHOOL ALGEBRA. 


(1) Find the cube root of a?+3a7b + 3ab? + 8b’. 


a + 3a7b + 3ab?+4+ B?|a +b 
3a? a? 


___+3ab +P? 3a2b + 3ab? + 68 
3a?4+ 3ab + Bb? 3a7b + 3ab? + b 


The first term a of the root is obviously the cube root of the first 
term a’ of the given expression. 

If a® be subtracted, the remainder is 3 ab + 3ab? + 6°; therefore, 
the second term 6 of the root is obtained by dividing the first term 
of this remainder by three times the square of a. 

Also, since 3a7b + 3ab? + b?= (3a? + 3ab + b7)b, the complete 
divisor is obtained by adding 3ad + 0? to the trial-divisor 3 a’. 


(2) Find the cube root of 82° + 36a°y + 54 x7?+ 277’. 


8a + 36 ay + 54 ay? + 2743 |2a + 3y 
12 2? 8 a3 


(6x+3y)3y= 18ay +9y?| 36a7y + 54 xy? + 27% 
1207+ 18ay+9y?| 360°%y + 54ay? + 2773 
The cube root of the first term is 22, and this is therefore the first 
term of the root. 
The second term of the root, 3 y, is obtained by dividing 36 zy by 
3 (22)? = 1227, which corresponds to 3a? in the typical form, and is 
completed by annexing to 12a? the expression 
{3(22)+3y33y = 18 ay + 9y?, 


which corresponds to 3ad + 0? in the typical form. 


225. The same method may be applied to longer expres- 
sions by considering a in the typical form 3a’+ 3ab+4 0° 
to represent at each stage of the process the part of the root 
already found.’ Thus, if the part of the root already found 
is x+y, then 3a? of the typical form will be represented 
by 3(a@+ y)’; and if the third term of the root be +z, the 
3ab-+ 6? will be represented by 8(a+y)z+2. So that 
the complete divisor, 3a?+ 3ab + 8”, will be represented by 
8(a+yl+3(e+y)z24+2. 


INVOLUTION AND EVOLUTION. ZirS 
Find the cube root of 2°— 32°+ 523—8zx—1. 


|e?—-a—1 


2 — 3 +522 — 3e—1 








(8a?—a2)(—2)= —38a3 + 

3a* — 3284 

— 32+ 623—32-—1 
3 (a? — 2)? = 3at— 623 + 3.2? 

(3 22—3a—1)(—l)= —3a7 +32 +1 

+3e +1|/—32'+622—32—-—1 





The root is placed above the given expression for convenience of 
arrangement. , 

The first term of the root, x?, is obtained by taking the cube root 
of the first term of the given expression; and the first trial-divisor, 
3x‘, is obtained by taking three times the square of this term. 

The first complete divisor is found by annexing to the trial-divisor 
(3 a? — x)(— a), which expression corresponds to (3a + 6)b in the 
typical form, ; 

The part of the root already found (a) is now represented by x? — «a; 
therefore 3a? is represented by 3(«?—a)?=3a4—6 25 + 327, the second 
trial-divisor; and (a+ 6)b by (8a?—3x2 —1)(—1); therefore, in the 
second complete divisor, 3a? + (3a + b)b is represented by 


(3 at —6 28 + 327) + (—3a?— 34—1)x(—-1)= 3at— 6034 3241. 


Exercise 73. 
Find the cube root of 
1. a + 8077 + 8az?+ x’. 
2. 841274 62'+ 2’. 
3. x°—8a2°-+ 5a’2’* — 8a°a — a®. 
4. 1—62+ 212’— 442°+ 638 2*— 542° + 27 x°. 
5. 1—824+ 62? — 7T2’*+ 62*— 82° + 2°. 
6. 2+1—6x—62°4+ 152°+ 152*— 202%. 


214 SCHOOL ALGEBRA. 


7. 64a°—1442°+ 8— 86x + 1022’— 171 a°4+ 204 2%. 
8. 27a°®— 27a°— 18a*+ 17a°+ 6¢?— 38a — 1. 

9. 82°— 362° -+ 662‘ — 682° + 8382°7—92+1. 
10. 27+ 108z2-+ 902’ — 802° — 60 2* + 482°— 82° 


Leeper aig 
LES. See aeceyy cae ser a Ree ort 


226. Arithmetical Cube Roots. In extracting the cube root 
of a number expressed by figures, the first step is to mark 
it off into periods. 

Since 1 =1’, 1000 =10*, 1,000,000 = 100%, and so on, it 
follows that the cube root of any number between 1 and 
1000, that is, of any number which has one, two, or three 
figures, is a number of one figure; and that the cube root 
of any number between 1000 and 1,000,000, that is, of any 
number which has four, five, or sex figures, is a number of 
two figures; and so on. 

If, therefore, an integral cube number be divided into 
groups of three figures each, from right to left, the number 
of figures in the root will be equal to the number of groups. 
The last group to the left may consist of one, two, or three 
figures. 


227. If the cube root of a number have decimal places, 
the number itself will have three times as many. ‘Thus, if 
0.11 be the cube root of a number, the number is 0.11 x 0.11 
x 0.11 = 0.001331. Hence, if a given number contain a 
decimal, we divide the figures of the number into groups 
of three figures each, by beginning at the decimal point 
and marking toward the left for the integral number, and 


INVOLUTION AND EVOLUTION. OT 


toward the right for the decimal. We must be careful to 
have the last group on the right of the decimal point con- 
tain three figures, annexing ciphers when necessary. 


| 228. Notice that if @ denotes the first term, and 6 the 
second term of the root, the first complete divisor is 
38a + 8ab-+ 0’, 
and the second trial-diisor is 3(a + 6)’, that is, 
3a+ 6ab+ 30’, 
which may be obtained by adding to the preceding complete 
divisor its second term and twice ils thurd term. 
Ex. Extract the cube root of 5 to five places of decimals. 
5.000 (1.70997 












1 
3 x 10? = 300 4000 
3 (10 x 7) = 210 
7 49 
559 3913 
259 87000000 
3 x 17002 = 8670000 
3(1700 x9) = 45900 
Q? 81 
8715981 78443829 
45981 85561710 





3 x 1709? = 8762043 78858387 
67033230 


61334301 


After the first two figures of the root are found, the next trial- 
divisor is obtained by bringing down the sum of the 210 and 49 
obtained in completing the preceding divisor ; then adding the three 
lines connected by the brace, and annexing two ciphers to the result. 

The last two figures of the root are found by division. The rule. 
in such cases is, that two less than the number of figures already 
obtained may be found without error by division, the divisor being 
three times the square of the part of the root already found. 


216 SCHOOL ALGEBRA. 


Exercise 74. 
Find the cube root of 
1. 4918. 3. 1404928. 5. 385828.352. 
2. 42875. 4. 127263527. 6. 1888.265625. 
Find to four decimal places the cube root of 
TR1. 9. 3.02. 11. 0.05: 13. 2. 15. 5%. 
8. 10. 10. 2.05. 12. 0.677. 14, 3. 16. +. 


229. Since the fourth power is the square of the square, 
and the sixth power the square of the cube, the fourth root 
is the square root of the square root, and the seath root is 
the cube root of the square root. In like manner, the 
eighth, ninth, twelfth, ...... roots may be found. 


Exercise 75. 
Find the fourth root of 
1. 8la*+ 1082? + 5427+ 122-41. 
162° — 82.aa0° + 2407s? — Saez at 


38. lt4e7+a°+42'+102°+162?+102?+ 192+ 162°. - 


wo 


Find the sixth root of 
fab dai bd’ Hib 20d" logs 
. 129—14582+12152?—5402°+ 185 2*—-182°+ 2°. 
. 1—18y+4+1357 — 5407? + 1215 44 — 1458 + 72944. 


ca Oo 


CHAPTER XV. 
THEORY OF EXPONENTS. 


230. If nm is a positive integer, we have defined a” to 
mean the product obtained by taking a as a factor m times. 
Thus a? stands foraxaxXa; 6 stands for bxXbxb~x 6. 


231. From this definition we have obtained the following 
laws for positive and integral exponents : 
beatocat ara 
ae te. 
jaue fe rf) uenebe at ys Jee tr: 8 
a 


TN. Van = a, 
Voge n a eae 


232. Since by the definition of a” the exponent denotes 
simply repetitions of a asa factor, such expressions as a3 and 
a~* have no meaning whatever. It is found convenient, 
however, to extend the meaning of a” so as to include 
fractional and negative values of n. 


233. If we do not define the meaning of a" when 7 is 
a fraction or negative, but require that the meaning of a” 
must in.all cases be such that the fundamental index law 
shall always hold true, namely, 

qm x q” — gore: 
we shall find that this condition alone will be sufficient to 
define the meaning of a” for all cases, 


218 SCHOOL ALGEBRA. 


234, To find the Meaning of a Fractional Exponent. 


Assuming the index law to hold true for fractional expo- 
nents, we have 


1 1 141 2 
a? xX a*=a? 4=4* = 4, 


ih an 1 
at alate atet gts 








atxatxatxat=attitita 
gt isd 1%. 
GX Onin to ” factors = an 


m m m sm mn 
—.....to nm terms 


an X Qn .+- to n factors =an*n 


n 
i fe ames 
a, 


| 


Sth 6 a 


That is, a? is one of the two equal factors of a, 
a is one of the three equal factors of a, 


ins 

a‘ is one of the four equal factors of a’, 
1 

an is one of the n equal factors of a, 


m 
an is one of the n equal factors of a”. 
1 = tere 
Hence 9 a8 Vat nase ey OF 


at = AV oy om (§ 213) 


Also, am» Xan X anXerw to m factors. 


m 
mae, Bea Tha += chee to m terms —=aqn. 


an (Va). 


The meaning, therefore, of ae where m and 7 are posi- 
tive integers, is, the mth root of the mth power of a, or the 
mth power of the nth root of a. 

Hence the numerator of a fractional ent indicates 
a power, and the denominator a root; and the result is the 
same when we first extract the root and raise this-root to 
the required power, as when we first find the power and 
extract the required root of this power. 


THEORY OF EXPONENTS. 219 


235, To find the Meaning of a’. 
By the index law, 
Ge Ca tee 
nal tat eg, 


*. a® = 1, whatever the value of a is. 


236. To find the Meaning of a Negative Exponent. 


If m stands for a positive integer, or a positive fraction, 
we have by the index law, 
ene Oat C, 
But Gee 
2x a I. 


That is, a and a~™ are reciprocals of each other (§ 167), 
po. thatia* — sa and aha. 
a 
237. Hence, we can change any factor from the numerator 
of a fraction to the denominator, or from the denominator 
to the numerator, provided we change the sign of its exponent. 


Thus 22 J 


cd® Gove: 





may be written al’c*d~, or 


238. We have now assigned definite meanings to frac- 
tional and negative exponents, meanings obtained by 
subjecting them to the fundamental index law of positive 
integral exponents; and we will now show that Law II., 
namely, (a”)"=a"", which has been established for positive 
integral exponents, holds true for fractional and negative 
exponents. 


(1) Ifn is a positive integer, whatever the value of m, 


We have (a™)" = a™ x a™ X a™..... to n factors, 


220 SCHOOL ALGEBRA. 


(2) If nis a positive fraction 2, where p and q are posi- 
tive integers, we have q 3 


P ee 
(am)n = (a™)t = V(am)P @ 234 
= Vane (1) 
mp 
=ad 3 234° 
am q™ x - 
= gmn 


(3) If nis a negative integer, and equal to —p, we have 
1 


(a™)" = (a™) "P= (amp 2 236 
1 
~ am @) 
=a-™ 2 236 
= qm(—p) 
=am, 


(4) If mis negative and equal to the fraction etm where 
p and q are positive integers, we have s 
i 











p 
(am)» = (a")~g= — 2 236 
(a™)¢ 
= oe 2 234 
a™ 
= — 3 234 
ag 
3 1 
q™ Xe 
ee! 
“ie qm(—n) 
q 
ay qm (1) 
= gn, 3 236 


Hence, (a”)" = a™", for all values of mand n, 


THEORY OF EXPONENTS. vival | 


239. In like manner it may be shown that all the index 
laws of positive integral exponents apply also to fractional, 
and negative, exponents. We will now give some examples. 








& | 1 1 
ibs TEN alae 
(2) 164 =(V16=+2=+8. 
(3) at x a7§ aati att = Va, 
(4) a aca t 2 ptt Pe gt 1 
(5) ata) at =e 
6 ee 
a 

(7) Vaib ed = athe tas. 

goes 1 41 1 a 
8 4 3)" 2 — = Se SS 
(8) ( sae (4a73)3 Ait 8a) T8 














(9) 


Wea\o* 2 / Bley) 275? Q7a%b* 
81 0° l6a* os Ny 8 


u ee a 





(10) (Baty = 





(GRE a g3x$-2 9t W/3 





Exercise 76. 
Express with fractional exponents : 
1. Va 3. Val. ob. Va 7. Va +V24+V168* 
2.Va. 4. V—8. 6 Va. 8. Vala + Vare. 


pe SCHOOL ALGEBRA. 


Express with root-signs : 








3 ro at ft Pat 
Das. Lie a10*; 1g ee he 15. a?— gz ¢5, 
2 ee oyu Rites tee 
10. 3. 12. a> b3, 14. 38a%y #. 16. a+ 2¢3. 
Express with positive exponents: 
—1,2 
Wier ar a 19.0 ae ae Ae Ue eee SOReE 
Sans 
18. a? 20. 4ay-7. 22. Ba-*b?. 
Write without denominators : 
2 
2 yo —2 1/3 
yee ae 26) y eee 28) ee ee 
Boe Oe One oa Ves trae 
2 DS ee ee | ne Fishin 6 
an, 22. Gy ead a gg. te: Pees 
= a-2b-2.c-4 an" b-5e-3 
Find the value of 
5 2 3 1 5 
30. 8°. Bo7 248: 34. 362. 36. (— 27) x 257. 


g1.16-4) 33)(—8) £38. (ee sia ol ne. 


Simplify : 
38. 88x 4-2, 40. (gts)? x (ga)P. 42. (a 2B)7 7, 
39. (Ax)?x 16-4. 41. a *bi x ath? 48. (a 282). 


leo 4°56 =2)¢= 1,"indithevalucior 
44. aid. 46. a 33% 48. 3 (ab). «0. (ade)? 


45. ab~’. 47. ba Fee 5 49. 2(ab)~*. 51. (ab e)*. 


THEORY OF EXPONENTS. 220 


240. Compound expressions are multiplied and divided as 
follows : 
fs 1 1 1 ab ae 1 
(1) Multiply «? + 2? yt +y*? by 247 — ay? +y?. 
wt a at yt “Lk y? 
+ vty? +f at ya ey 


x + why? +y 
(2) Divide V2?+ Vx—12 by Vr—8. 
at+ xt —12|2t —3 





at — 322 at +4 
+43 —12 
+ 40% — 12 





Exercise 77. 


Multiply : 
1. a+? by at — 82. 3. a? — bt by at — 5. 
ag at + b2 by at + B?, 4. ee +2 by a? — Qe. 


bit poy te by eee ay 

6. ot —xtytt ys by ot yp 

ie ateyty by ot — yf. 

8. 1+0'+5°? by 1—b*+0°. 

9. attatyt ty? by at—atyt ty’. 

10. ab? +42at —3! by 28-4 4a-* —6a- #8}. 


224 SCHOOL ALGEBRA. 


Divide : 
11. a—b by as — 83, 13. a—b by at — bt, 
12. a+6 by at +. 8. 14. a+ by ak + BS. 


15. Qa?%4+ 6a2-1y1— 162°y“ by 224 2ey1+ 4x y~, 

16. ety+tz—B8aty 2 by ot + yh 4 28. 

17. «--8254+82?—1 by 28-1. 

18. stotytty by ot — at yt + yf? 

19. ef? —4e3+146a°2 by a? — 2, 

20. 9 —122? 2449724 23 by 8a? —2— 273, 
Find the square root of 

21. 28420341. 23. vi—4x34 4, 

22. 4a8— dab bt+ 2. 24. 4a*4 40741, 

25. 9a—12a?+10--4a 24a". 

26. 49a? - 282+ 18a? 42341. 

27. m'-+-2m— 1—2m7)+- m-. 

98. 14+ 4y°§—2y°8— 424 257 $2478 4 169. 
Expand : 

29. (e—ax). 31. (2al—a-}). 33. (4 Ve—4 Vay. 


380. (Ve—hx). 32. (Qe%+a8) 34. (AV eF4+427)) 


CHAPTER TA VI. 


RADICAL EXPRESSIONS. 


241, A radical expression is an expression affected with 


the radical sign; as, Vu, V9, Va, Va-+, V/B2. 


242. An indicated root that cannot be exactly obtained 
is called a surd, or irrational number. An indicated root that 
can be exactly obtained is said to have the form of a surd. 

The required root shows the order of a surd; and surds 
are named quadratic, cubic, biquadratic, according as the 
second, third, or fourth roots are required. 

The product of a rational factor and a surd factor is 
called a mixed gurd; as, 83V2, bVa. The rational factor 
of a mixed surd is called the coefficient of the radical. 

When there is no rational factor outside of the radical 
sign, that is, when the coefficient is 1, the surd is said to be 


entire; as, V2, Va. 


243. A surd is in its semplest form when the expression 
under the radical sign is integral and as small as possible. 

Surds which, when reduced to the simplést form, have 
the same surd factor, are said to be similar. 


Nore. In operations with surds, arithmetical numbers contained 
in the surds should be expressed in their prime factors. 


REDUCTION OF RADICALS. 


244, To reduce a radical is to change its form without 
changing its value. 


226 SCHOOL ALGEBRA. 


Case I. 


245. When the radical is a perfect power and has for an 
exponent a factor of the index of the root. 


(1) Va = at=a?= Va; 
(2) 3602s? = V(6.ab)'= (6ab)% = (6ab)? = V6ab; 
(3) V2Baib'ee = WV (Sabb!) = (Sarbet)é = (Barbet)s 
= VBavbe'. 
We have, therefore, the following rule: 
Duwide the index of the root by the exponent of the power. 


Exercise 78. 





Simplify : 
eh ea 6. Var. a Reo 
2. 16. Vent Cu sd 
3. V27. 8. Vaib'. 12 eee 
4, /49. 9 rey 





Be OH te ee 
tie 13. AES 
5. V64. 10. V16a‘bt. 

Case IT. 


246, When the radical is the product of two factors, one of 
which is a perfect power of the same degree as the radical. 


Since Va" = Va" x Vb =aV0 (§ 215), we have 
(AIS Ee ein) GN! be 
(2) V108 = V27 x 4 == V27 x V4 =38V4; 


RADICAL EXPRESSIONS. OF. 





(3) 4V 7202 = 4-V36 0b? xX 26 = 4V36 0b? x V26 
=4 x 6abWV2b = 240d V286; 

(4) 2V 540% = 2V 27 a? X 2ab = 2V 27a xX V2ab 
=2x 8av2ab =6a V2ab. 


We have, therefore, the following rule: 


Resolve the radical into two factors, one of which is the 
greatest perfect power of the same degree as the radical. 

Remove this factor from under the radical sign, extract 
the required root, and multiply the coefficient by the root 
obtained. 


Exercise 79. 

















Simplify : 

V V28. 1S emi 144 on. of G4) 
BT o. 14. 8V min. eRe: 
3. iV 12. 15. 3V2Fa°, 26. ee 

4. ~/500. 16. 2V act. a 

5. W432. 17. Un/arpe, T ARS 
6. V192. 18. 7V8a°d. Rue 
ra Na 

8. W243. 20. 4V aly’. Sa. eee 

9. V176. Bie 1020, Vi296 
10. W405. 22), Ni 187. 30. Jee. 
Tien! 110, 23. »/1250. ae 
12. 3/864. 24. 41/648. 31. 80 N9a%b? 


928 SCHOOL ALGEBRA. 


e Case III. 


247. When the radical expression is a fraction, the denomi- 


nator of which is not a perfect power of the same degree as the 
radical, 


oe a lee nee 
x Sie, aloe 7 
: OY SC = Sat DAS 
Vi- eat 4x9 * 36 RY 
A eh gs 315 xX BxX 4 .| if 
DOs Ae Ota 0 aaa 
V2- Ves 87 x8 87 x 8 


1 mane 
= 3579 V60 =} V60. 








We have, therefore, the following rule: 


Multiply both terms of the fraction by such a number as 
will make the denominator a perfect power of the same 
degree as the radical; and then proceed as in Case LI. 


Exercise 80. 














Simplify : 
Lay a 4. 74. Tie 10. 2V3. 
2. 3/2. 5. W258. 8. V2. 11. 8V x. 
AES 6. 8V2. Q fain) eee 12, oa) 2, 
ate 3! aa (o# 
1D.iq tees 1B ies, 17, cm 
BF B ba? 
4 ; OD 7322, 
14. 5 16. es, 18, 2x) ae 
a 125% Bxuye* 


RADICAL EXPRESSIONS. 


b9 
iw) 
We) 


Case IV. 
248. To reduce a mixed surd to an entire surd. 
Since aVb = Va" x Wb = Va"b, we have 
(1) 85 = One a Oo ey 46 
(2) abVbe = V (ab) x be = Vail? x be = Va'b'c ; 
(8) 2aVay = VOa x ay = BPX ay = VB: 
(4) 87Vei=VBy/) xX = VBL ye. 





We have, therefore, the following rule: 


Raise the coefficient to a power of the same degree as the 
radical, multiply this power by the gwen surd factor, and 
mdicate the required root of the product. 


Exercise 81. 


Express as entire surds: 
1. 5V5. iy NERS ON ote Sse LAL 
Mele 68 BY 2.8 10.5 OV). - 14804 
oe a ao ae LL eer! 10 1b rdw gant 
4. 2/4, Sb eN/ 4 12 oa 16. —4Vm'. 


Case. V. 
249, To reduce radicals to a common index. 


(1) Reduce V2 and V3 to a common index. 
V2 =a 28 - V8 = V8. 
Agee Ae a Ok afr 8/9: wpe Et eye: 


~ Hence, 


230 SCHOOL ALGEBRA. 


Write the radicals with fractional exponents, and change 
these fractional exponents to equivalent exponents having the 
least common denominator. Raise each radical to the power 
denoted by the numerator, and indicate the root denoted by 
the common denominator. 


Exercise 82. 


Reduce to surds of the same order : 


1. V3 and V5. 7. V2, V3, and V5. 

2. V14 and V6. 8. Va?, Vb, and Ve. 

3. VW2and V4. 9. Vat, Ve, and V2. 

4, Vaand Ve. 10. Vay, Vabe, and V2z. 
5. V5 and V5. ll. Vz—yand Vz+y. 
6. of of and 2%, 12. Va+tb and Va—b. 


Nore. Surds of different orders may be reduced to surds of the 
same order and then compared in respect to magnitude. 


Arrange in order of magnitude : 
13. V15 and V6. 15. 80, V9, and V8. 
14. V4 and V3. 16. V3, V5, and V7. 


ADDITION AND SUBTRACTION OF’ RADICALS. 


250. In the addition of surds, each surd must be reduced 
to its simplest form; and, if the resulting surds are similar, 


Find the algebraic sum of the coefficients, and to this sum 
annex the common surd factor. — 
If the resulting surds are not similar, 


Connect them with their proper signs. 


RADICAL EXPRESSIONS. at 


(1) Simplify V27 + -V48 + -V147. 
V27 = (3? x 3)? =3 x 38 =3V3; 
V48 = (2 x 3)8 = 2 x 88 = 4 x 38 =4V3; 
VI47 = (7? x 3/2 = 7 X 88 = 7 V3. 
 V27 + V48 +. V147 = (3 +44 7)V3 =14V3. Ans. 


(2) Simplify 27320 — 340. 
2V320 = 2(28 x 5)P = 2x 2x BE = 8 V5; 
PNG ard hg AM ata 
. 2/320 — 3/40 = (8 —6)V5=2V5. Ans. 


(3) Simplify 2V3—8-V2 + V+. 


2V3 = 2V18 =2V15 x t= 2Vv 15; 
3V3 =3V15 =38v15 x = 3V 15; 


— 4x 15 ys x + 4 
Vis=¥ i? 16x = a vib, 


o. 2V8—3V84 Va = (2-244) VIB =4 V5. Ans. 





Exercise 83. 

Simplify : 

LN Yea bie BAA 

8. 2/73 —5V/3 + 9v3. 
. V4 2782-7108. 7. VO7+ W484 V5. 
. BV2+4 412 — V64. 8. 4V147+3V754+ V192. 
i AV5 + 24-V5 + 4V40. 9. Vativat+ va. 
. BYV8— 548-2 4/243, 10.. Va? + £V ai — BV 2703, 


Yow SCHOOL ALGEBRA. 


11a + ba BN: \ 
12. V25b+ 2V9b — 8V46. 

13. 2V115 — 3-63 + 5-V28. 

14. V3+48V32+1V128 — 6V18. 

15. V75+ V48 — V147 + V300. 

16. 20V/245 — V5 + V125 — 21-180. 

17. 2V20 +4V12 — 2V27 + 5V45 — 9-V12. 
18,0 7/254. 41/45 —4/9 — 9N/80- / 20-4 64. 
19. V54+ Vi — 7250 — 8v2. 

20. 2V$4+ V60—-V15-+ V84 V8. 

21/87 A/S Ee 

22. Wah — Vb + W326. 

23. Vata + Voix — V4 0°b'x. 

24. W4ar2 + Vyie+ Vatz. 

25. Va'b’ec—av4e+bvare. 

26. W8la®— V16a+ V256a8. 

a7. V27 mt — 125m + V216m. 

28. V8a—V50a' — 38-V 18a. 

29. 6av63ab)— 8-V/112.0°8? + 2.ab-V343 ab. 
30. 8-V 125 mint + nV 20m —-V500 mini. - 
haya V382.0'0' + 6V'726 + 8V1280°d. 

32. 2%/a% —3a?X/646 + bav/a'b + 20?/ 1250. 


22. 
23. 
24. 
25. 


RADICAL EXPRESSIONS. 


233 


MULTIPLICATION OF RADICALS. 


251, Since Va x Vb = Vab, we have 
(Lr BV/8 ons So Bey BG 9 == 15/16 = 60; 
(2) 8V2 x 4V38 =8VB x 4V9 = 12-772. 

We have, therefore, the following rule: 


Express the radicals with a common index. Find the 
product of the coefficients for the required coefficient, and the 
product of the surd factors for the required surd factor. 


Reduce the result to its semplest form. 


VENI S x OF 


. VbxX V 20. 
MN NS 


.V38xXVv9. 


. V2x V2. 


VOIX V3. 


4 


8 


9. 


10 


11 


12 


Exercise 84. 


. W4EX VB. 
ae. 
Von! 
MN 8 V0 
. VW3xVI18. 
aN 6GN/ 8, 


13 


14. 


15. 


. W54 x V9. 
OV8 x V2. 
8X Wea 4. 


BATT Sein = 49. 
. V81xV—45. 


. £V18 x 8V3. 


19. (VIB + 2V72—3-V8) x V2. 
20. (W/32—4V864 + 3Y4) x V3. 
Be oT LON 8 T4828) V8, 


V5 x W4. 


V16 x 4/250. 
64 x /16. 
V3 x W72. 


26. 


eo. 


27. 


28. 


ee 
Vex V4. 


V81 x V3. 
Vax Vi. 


30 


31 


32 


33. 


: Va xXV4. 
. V22x Ve. 


. Vy x V8. 


WV7x V5. 


234 SCHOOL ALGEBRA. 
252. Compound radicals are multiplied as follows: 


Ex. Multiply 2V34+8~Vz by 8V38—4V<z. 
2V3 4+3Vx 
3V3 —4V x 
18 +9V3a" 
—8V32—12e 


18+ vV3e—122 


Exercise 85. 


Multiply : 
1. VB+V4by V5—V4 4. 84+8-V2 by 2— V2. 
2. V9—Vi7by V9+VI17. 5. 54+ 2V3 by 83—5V3. 
3. 34 2V5 by 2— V5. 6. 3— V6 by 6—3-V6. 
TaN © On bby vB Oy a} 

Be V/A yeaa el 

9. V9 — 2/4 by 40/3 + V2. 

10. 2V30= 3/5 +578 by V82 v3 — V5, 

1135 = 85/3 Lat by Bd/Ba b ve. 

12. 4V8+4-V12 —1V32 by 8V32 —4V/50 — 2-V2. 

18. V6—V3-+ VIG by V36 + V9— V4. 

14, 2g -8V$ + 38V% by 8V2— V12— V6. 

15. Ive - 4/3 — V5 by 83V% — 5-V30 — 2V18. 

16. 2V18 + 8V3 + 6V4 by 2V13-+ 3VB + 6V§. 


RADICAL EXPRESSIONS. 235 


DIVISION oF RADICALS. 


258. Since Mab — Vax V6 = Vb, we have 


Va Va 
Qiu Se ee 
2/3 | 
(2) 4V/3_4V3? _4V3? x OF \/79. 


2V2 IVR ONY 88 
We have, therefore, the following rule: 


Express the radicals with a common index. Find the 
quotient of the coefficients for the required coefficient, and the 
quotient of the surd factors for the required surd factor. 

Feduce the result to its sumplest form. 


Exercise 86. 


Divide: 

det aaa om dia \/ eeby AN) ite Vi by V24. 

2. VBI by V3. 5. VE by V3. 8. VEER by VE. 

SV oau byV a. » 6. V4 by VE. 9. VSP by Vit 
10. 8V6-+ 45-V2 by 8-V3. 

Pi os/ 5308 by ov 15. 

12. 84V15+168V6 by 8V21. 

13. 30V4—386V/10+ 30V90 by 3-20. 

14. 507184 18V20—48V5 by 2V380. 
t5..V54 by V386. 17. VI2 by V6. 19. V¥ by V33. 
16. V49 by V7. ‘18. V8 by V6E. 20. V2x by V x4. 
a1. ¥0064 by VIO. 22. VF = by w+y.- 


236 SCHOOL ALGEBRA. 


254, The quotient of one surd by another may be found 
by rationalizing the divisor ; 


that is, by multiplying the 


dividend and divisor by a factor which will free the divisor 
of surds. 


255. This method is of great utility when we wish to 


find the approximate numerical value of the quotient of 
two simple surds, and is the method required when the 
divisor is a compound surd. 


ne 


(1) Divide 8V8 by V6. 


BVB_6v2_evixv8_6VB_ pg 
V6 v6 vV6xv6 Ge tan: 


(2) Divide 8V5—4V2 by 2V5 + 38-—V2. 
8V5—4V2_ (38V5—4V2)(2V5—3V2)_54-17V10 
2V5+3V2 (2V54+38V2)(2V5—8vV2) 20-18 


_ 54—17V10 _ oy _ 81/10. 


Z 
5 5 
(3) Given /2 = 1.41421, find the value of We 
SR RE NE) 


MEAN 8 Me 
Ree: Exercise 87. 


Va +Vb ‘by Vab. 7. 8+5V7 by 8—5V7. 


V125 by 5V65. Bi Bin/B by 4./ 8 8/2. 
8 by 114+38-V7. 9. T5V14 by 8V2+4 2V7. 
8V2—1 by 8V2+1. 10. V5—V3 by V5+-V3. 


17 by 8V7+2V3. 11. V82 V7 by V7 V2, 
Liby V2 dan 8: 12. == BAAL0U by sb dn/5. 


RADICAL EXPRESSIONS. sith 


Given V2—1.41421, V3=1.73205, W/5 = 2.23607; 
find to four places of decimals the value of 





He, ae 16. ey 19. ioe 22; T= 38Vv5 
V2 500 8V2 5 +475 
iy | a ep pelos 20. ae 23. B+ V5 
V3 243 125 x/5 — 2 
Tenens, AP gal di eke oa, 3V2—1 
V5 ONS 4/5 Bn/2 


INVOLUTION AND EvouuTIon oF RADICALS. 


256. Any power or root of a radical is easily found by 
using fractional exponents. 


(1) Find the square of 2Va. 
(2Va)? = (2.08)? = 2 a8 = 4a? = 4 Va. 
(2) Find the cube of 2-Va. 
(2Va)3 = (2.03) = 23 a? = 8a? = 8 ava. 
(3) Find the square root of 4a-Va*b’. 
(4. 2-V 0263)? = (4 xa2b2)3 = 42 otah dt — 43 talked — 2Var8x?. 


(4) Find the cube root of 4a-Va'd’. 
(4.0-Va3b3)3 = (4 wa2h?)8 = 44 oh a3bt = 46 wtadht — V16 a3d82?, 


Exercise 88. 


Perform the operations indicated : 
= hor (d bpapatcen erent 
1. (Wnt. 3. (Wat)¥, 5. VV (0—y 
: 6) Fone NSE 
Va — b)*, 


se 


2. (vm), 4. (Wy). 


238 SCHOOL ALGEBRA. 

7. (V2ab). 10. VV. 13. Ban 2b*, 
8. (Ve—yy. 11. Vr/729. 14, 2/3208, 

9. (Wa)t. 12. VV 135. 15. 4/1289/248 a, 





PROPERTIES OF QUADRATIC SURDS. 


257. The product or quotient of two dissemilar quadratic 
surds will be a quadratic surd. Thus, 


Wie RRS to 
Vabe+ Vab = Ve. 

For every quadratic surd, when simplified, will have 
under the radical sign one or more factors raised only to 
the first power; and two surds-which are dissimilar cannot 
have ad/ these factors alike. 

Hence, their product or quotient will have at least one 
factor raised only to the first power, and will therefore be 
a surd. 


258. The sum or difference of two dissimilar quadratic 
surds cannot be a rational number, nor can it be expressed 
as a single surd. 

For if Va+-Vé could equal a rational number ec, we 
should have, by squaring, 


at+2Vabtb=c'; 
that is, +%VJVab=e—a—b. 
Now, as the right side of this equation is rational, the 
left side would be rational but, by $207; Vab cannot be 
rational. Therefore, Va+ Vé cannot be rational. 


In like manner, it may be shown that Va + Vb cannot 
be expressed as a single surd Ve. 


RADICAL EXPRESSIONS. 939 


259, A quadratic surd cannot equal the sum of a rational 
number and a surd. 


For if Va could equal ¢-++ Vb, we should have, by 
squaring, 
a=e+2cvVb+, 
and, by transposing, 
Dinh ine b= ee 
That is, a surd equal to a rational number, which is 
impossible. 


260. If at+Vb=2+ Vy, then a will equal x, and b 
will equal y. my 

For, by transposing, Vb— Vy=2—a; and if 5 were 
not equal to y, the difference of two unequal surds would 
be rational, which by § 258 is impossible. 

Diet MONCUO) ee: 

In like manner, if a— Vo6=2—-Vy, a will equal 2, 

and 6 will equal y. 


261. To extract the square root of a binomial surd. 
Ex. Extract the square root of a+ Vb. 


Suppose Vat Vb = Va + Vy. (1) 
By squaring, a+ Vbs=2+4+2Vay + y. Cl | 
“ @=n2+y and Vb=2Vay. 3 260 
Therefore, a—Vb=x2-—2Vay+y, (3) 
and Va~Vb= Vi~ Vy. (4) 
Multiplying (1) by (4), 
Vat—b=2-y. 


But a2=2+y. 


240 SCHOOL ALGEBRA. 


Adding, and dividing by 2, «={**+¥4— vend 
‘Subtracting, and dividing by 2, 
_a—Va'—b 


¥ 


: ! 
: _\% tVa?—~b  ala— Va — 6. 
ee Va + Wb Viet so tn + 5 


From these two values of # and y, it is evident that this 
method is practicable only when a?-- 6d is a perfect square. 





(1) Extract the square root of 7 eAn/8. 


Let Vi + Vy = M74 4V3. 

Then Vi — Vy = V7—4v3. 

Multiplying, xe—y = V49 — 48. 
 &—y=, 

But ; e+y=7. 


* o=4, and y=3, 
<a gene a 
V7 44 V3 = 24 V3. 
A root may often be obtained by inspection. For this purpose, 


write the given expression in the form a + 2V, and determine what 
two numbers have their sum equal to a, and their product equal to b. 


(2) Find by inspection the square root of 75 —12V21. 


It is necessary that the coefficient of the surd be 2; therefore, 
75 — 12V21 must be put in the form 


75 — 2V'756. 


The two numbers whose sum is 75 and whose product 1s 756 are 
63 and 12. 


Then 75 — 2V'756 = 63 + 12 — 2V63 x 12, 
= (V63 — V12). 
That 18, V63 — V12= square root of 75 — 12V21, 


or, 3V7 —2V3 = square root of 75 — 12V/21. 


RADICAL EXPRESSIONS. 241 


Exercise 89. 


Find the square root of 


Pep ANB. OG Wis PGA AONE 
2a N78: Sey yee oN / Oe 14 ye be ON oO. 
Shr ON 10: 9. 19+ 8V3. a5) AT 44/88) 


4. 18+8~V5. 10. 8V6-+ 20. 16. 29+ 6/22. 
5. 842/15. pel shee av AD 17. 83+ 12/35, 
Cre toe 44/14)" 12n bl 86V2. 18. 55 — 12/21. 


EQUATIONS CONTAINING RADICALS. 


262. An equation containing a single radical may be 
solved by arranging the terms so as to have the radical 
alone on one side, and then raising both sides to a power 
corresponding to the order of the radical. 


Ex. V2#?7—9+2=9. 
era Spas 
By squaring, 2? — 9 = 81 — 182 + 9. 
182 = 90. 
od Meas), 


263. If two radicals are involved, two steps may be 
necessary. 


Ex. Vze+15+~V2c=1. 


Vat 15 + Ve = 15. 
Squaring and simplifying, we have 
Vx? + 152 = 105-2, 
Squaring, we have x? + 15a = 11025 — 2102 + a3, 
225 7 = 11025. 
. © = 49. 


242 


1 


a 


3 


4. 


5. 


19. 


20. 


SCHOOL ALGEBRA. 


Exercise 90. 


Solve: 
IV +5 = V8. §. 5/82 a =30. 
S\/4 ge 8 SAV 1B ee Be vel 4 A 4p Sed) 10. 


Va +9 = 5V2 —8, 10. Vl0y—4=V7y+ 11. 


4 = IVa — 8. 11. 2V 2% —2—V82 (a = 2), 
eee. 12, Vip +a2=8+4-~V2. 
T+ 2V82=5. 13. V32+2=16—-V*z. 
JV2228 =+. 3, 14. Ve —-Va—5= V5. 


15.2 -- 90. V/s B= 0: 
16. Ve+15—-7=17 —V«— 18. 
Ll SNA a ae; 
18) Ne ee ee 
Ve—8 i Ven) gy ee aie 








Vet8 Ve 2 1— (2) 

Les V3 1 36 ii bes 3 
Bo ANG ADS a 22a = 38)F 4 ete 2: 
Ai a oan x (a—8)? 


23. Va+Ve+Va— Vex Ve. 
24. Var—-1=4+4Var—}. 
25. 8BVa —8Va = Va — Va+ 2Va. 


5 


2653/9 he aN Oe ee 
V9+2a 





CHAPTER XVII. 
IMAGINARY EXPRESSIONS. 


264. An imaginary expression is any expression which 
involves the indicated even root of a negative number. 

It will be shown hereafter that any indicated even root 
of a negative number may be made to assume a form which 
involves only an indicated square root of a negative num- 
ber. In considering imaginary expressions, we accordingly 
need consider only expressions which involve the indicated 
square roots of negative numbers. 

Imaginary expressions are also called imaginary numbers 
and complex numbers. In distinction from imaginary num- 
bers, all other numbers are called real numbers. 


265. Imaginary Square Roots, Ifa and 8 are both posi- 

tive, we have 
eNO een eNO Ree TL LCN) id, 

If one of the two numbers a and 0 is positive and the 
other negative, law I. is asswmed still to apply ; we have, 
accordingly : 

ai Py = NAR 1 ONT 

SE esi as Saas Naan 

Ra HN ET aT 
and so on. 

It appears, then, that every imaginary square root can 
be made to assume the form aV—1, where a is a real 
number. 


944 SCHOOL ALGEBRA. 


266. The symbol V— 1 is called the imaginary unit, and 
may be defined as an expression the square of which 1s — 1. 


Hence, V—1xX V—1= Cy Ay =—l1; 

V—ax V—b=Vax V—-1x VbxvV—1 
= Vax Vb x(V—1)} 
= Vab X(-1) 
ae aD, 

267. It will be useful to form the successive powers of 
the imaginary unit. 

yrds a aN aPC mere Spe tceeany fi Ps Mint oa EL 

SO AL) a ei RR ae BN eA ab ro Sl 

(V=1"=(V=1p V=1 = (- 1) V=i =- v1; 

(VST Vy St (See (Lae are 

(V=1) = (V1) V=1 = G1) v—l=4+ vel; 
and soon. We have, therefore, 

(Vat =4 VAT; 
(V— 1)? — oe 
(VE Stasis 
(V— 1)" = +1. 

268, Every imaginary expression may be made to assume 
the form a+6V—1, where a and 0 are real numbers, and 
may be integers, fractions, or surds. 

If 6=O, the expression consists of only the real part a, 
and is therefore real. 


If a=0, the expression consists of only the imaginary 
part 6\/— 1, and is called a pure imaginary, 


IMAGINARY EXPRESSIONS. 245 


269, The form a+~V—1 is the typical form of imaginary 
expressions, 

Reduce to the typical form 6+ V— 8. 

This may be written 6a) 1, Or.6 took: 
here a=6, andb=2V2. ~ 


270, Two expressions of the form a+6 4) SOT ee AY fla 
are called conjugate imaginaries. 

To find the sum and product of two conjugate imagi- 
naries, : 


at bn/— 1 
aeove 1 
The sum is 2a 
Gt DENA I 
a—b V—-1 
a tabVva1 
—abV—14+2 
The product is a + > 


From the above it appears that the swm and product of 
two conjugate imaginaries are both real. 


271. An imaginary expression cannot be equal to a real 
number. 


For, if possible, let 
at+bV/—l=e. 
Then transposing a, 6V—1=c—a, 
and squaring, —O = (¢— a)’. 


Since 4? and (e—a)’ are both positive, we have a nega- 
tive number equal to a positive number, which is impossible, 


946 SCHOOL ALGEBRA. 


272. Lf two maginary expressions are equal, the real parts 
are equal and the maginary parts are equal. 


For, let a+bV—1l=c+dv-1. 
Then (6—d)V—1l=c~—a; 
squaring, —(b—d)=(e—a)’, 


which is impossible unless 6 =d and a=ce. 


273. If x and y are real and x+yV—1=0, then x=0 
and y=0. 


For, UN a i ah 
[Site y tes 2, 
x + Se = O, 


which is true only when x=0 and y=0. 


274, Operations with Imaginaries. 


(1) Add 5 iV lLandeeeow ok 


The sum is Be BT Bee 1 
or Level! 


(2) Multiply 8-+ 2V—1 by 5 —4V—1. 
(35 2 1)(5 47-1) 
alba 12414 10 1 = 84) 
= 23 —2/— 1; 


(3) Divide 14+5V—1 by 2—38V—1. 

1445V—1_ (144+5V—1)(2+3V-1]) 
2 BN od (2 Bae ea) 

_13+52V=1 

4—(—9) 
_138+52V-1 
13 
=1+4V-1 


Leen) 9; 9. 
BN) =e16) 10. 
3. V—25. 11 
4, \/— 144, 12 
Bote 169: 13 
6. V— 27. 14 
Tay = 8" 15 
8. iV — 256. 16. 
Add: 
25. 
26. 
27. 
28. 
29. 
30. 
31. 
32. 
Multiply : 
33. V—8 by V—5. 


IMAGINARY EXPRESSIONS. O47 


Exercise 


Reduce to the form 6V—1: 





os ODA. 
vi 86, 


. V— 64. 
~ Va 729. 
. V— 289. 
. V—1024. 





Bon f a?) 


V/s 


91. 


Lee iia 

18. V—4. 

19. V— ath. 

20. V— 9xt. 

21 Vine (Qaapayye, 
DOIN eee Day) 
23. V—(a?+ 4). 
24. V— (a? — 7’). 


Se ae, PAL, 
Wen 64 ee Nn 36. 
WEI SEN, BEV OEY Yarn 
Nee ye a Na 
1 ls NN ee aN SE 
Doan hee Oe owen 1: 
Goby -al=bow ON 1, 
8aV—1—(2a—b)V—1. 


34, —V/—5 by V—5. 
$6216) by -/— 9. 


36.0 \/ a" bys 2a. 
BT ee yeN/ =n ye 
38. V—8 by V—16. 


248 SCHOOL ALGEBRA. 

BOo Vina 85 by Od. ato 28 by oye. 
40. V—(a+b) by Vt(a—6). 42. —5V—2 by 2V—5, 
ag) NS BE A / eS yy id 

ee Yves 
440 a, 
x 5 by x a 
45. avV—a+tbV—6b by aV—a—bv—6. 
46) 9\/ 2 1384/8 hy BV 4 — 2 Lo: 
47, W842V—3 by V8 —2V—8. 
48.) mo BV ob by n+4V—e. 


Perform the divisions indicated : 


























AG wie Bi eer, 61. try 
val V2 3 —V—2 
cee nae Ah sald gone eVene 
V— 6 V-2 Lay a 
a ey 57. Vow eg: Sevag 
ed ~V~ a—aV—1 
52. View 58. nas, 64. vb + Van 
V— 81 necks NAG aly ae 
53. v=a Bo! V— 1024 65. 2a+8bv—1 
Vie b V—5a F4—3bV —1 
54. Vaan 60. 8v— at 66. 0 PON ret 





CHAPTER XVIII. 
QUADRATIC EQUATIONS. 


275. We have already considered equations of the first 
degree in one or more unknowns. We pass now to the 
treatment of equations containing one or more unknowns 
to a degree not exceeding the second. An equation which 
contains the sguare of the unknown, but no higher power, 
is called a quadratic equation. 


276. A quadratic equation which involves but one un- 
known number can contain-only : 


(1) Terms involving the square of the unknown number. 
(2) Terms involving the first power of the unknown 
number. 


(3) Terms which do not involve the unknown number. 


Collecting similar terms, every quadratic equation can be 
made to assume the form 


ax’ + bz+e¢=0, 


where a, 6, and ¢ are known numbers, and z the unknown 
number. 

If a, 6, e are numbers expressed by figures, the equation 
is a numerical quadratic. If a, 6, ec are numbers represented 
wholly or in part by letters, the equation is a literal quadratic, 


277. In the equation az’?+ ba+c=0, a, b, and ¢ are 
called the coefficients of the equation. The third term ¢ is 
called the constant term, 


250 SCHOOL ALGEBRA. 


If the first power of x is wanting, the equation is a pure 
quadratic; in this case b= 0. 

If the first power of # is present, the equation is an 
affected or complete quadratic. 7 


PURE QUADRATIC EQUATIONS. 


278, Examples. 
(1) Solve the equation 52’— 48 = 2a”. 


We have 5 a? — 48 = 222, 
Collect the terms, 3 a2? = 48, 
Divide by 3, a? = 16, 
Extract the square root, e=+4, 


It will be observed that there are two roots, and that these are 
numerically equal, but of opposite signs. There can be only two 
roots, since any number has only two square roots. 

It may seem as though we ought to write the sign + before the x 
as well as before the 4. If we do this, we have +a2=+4, —x=—4, 
+e=—4 —v2=+4+4. 

From the first and second equations, s=4; from the third and 
fourth, «=—4; these values of x are both given by the equation 
a=+4, Hence it is unnecessary to write the + sign on both sides of 
the reduced equation. 


(2) Solve the equation 32’?— 15=0. 


We have og =won 
or Sea 0): 
Extract the square root, e=tvd, 


The roots cannot be found exactly, since the square root of 5 can- 
not be found exactly; it can, however, be determined approximately 
to any required degree of accuracy; for example, it lies between 
2.23606 and 2.23607. 


(3) Solve the equation 827+ 15=0. 
We have 3a7= — 15, | 


or a? = — 5, 
Extract the square root, g=+V—5, 


QUADRATIC EQUATIONS. 251 


There is no square root of a negative number, since the square of 
any number, positive or negative, is necessarily positive. 

The square root of —5 differs from the square root of +5 in that 
the latter can be found as accurately as we please, while the former 
cannot be found at all. 


279, A root which can be found exactly is called an exact 
or rational root. Such roots are either whole numbers or 
fractions. 

A root which is indicated but can be found only approx- 
imately is called a surd or irrational root. Such roots 
involve the roots of imperfect powers. 

Rational and surd roots are together called real roots. 

A root which is indicated but cannot be found, either 
exactly or approximately, is called an imaginary root. Such 
roots involve the even roots of negative numbers. 


Exercise 92. 











Solve: 

as ta Ce: 2 
1. 82°7—2=2'+6. 9. _ 2 + ora s, 
2. 527+-10=—627+- 1. 
3. Tx?— 50 = 407+ 25. 10, O& +8 li-# _, 
4. 62? —2=:427' +12. ; 

3 i| 7 

ek Pee ee, 
B. === 10, 4 6a? 8 
a8, wb Bat 
a ahaa igasine a 
1 plan Oe ae g—l] «+1 4 
2; A ema a ile wl 14. ala 7 a 





7 5 Sig | 2 Esa 


252 
15. 
16. 
17. 


18, 


ee 


20. 


21. 


22. 


23. 
24. 
25. 
29. 


30. 


280. Since (v7 + 6? = 2? + 2br7+4 0’, it is evident that 
the expression 2? + 26x lacks only the third term, 6°, of 


SCHOOL ALGEBRA. 
8027+ 1l2=10r¢+8+27+2. 
(x +4) (2 +5) =8(e¢4+1)(#+2) —4. 
3 (x —2)(@ + 8)=(#@+1)(e@+2)+2°+5. 
(Qe +1)(82—2)+ (l—2z)(84+42) = 827 — 15. 


7 fe Oy Dae? se Yh +10 

















—__— — ——___ +. —_—- = ]4, 
Li 9 5 
Bi rs yr eed ig el Oe 
: a 9 err 10s 
LO thee Gad Bak Ae Qios Car a8: 
18 lla2’?— 8 9 
alle ct At Py eile Oy en ee 
tt+loa—l1 2 hewn ed 
ax’?+b=e, 27 pak otek. somk ben! 








az? + b = ba? +a. 





28. . 
v+2br+e=b(22+1). x—b Ox 


25(a@+ a)(@+ 6)+ (e«x—a)(x# — b){=a+ 40. 


2$(x — a) (a+ 6) + (x 4+ a) (x — 6)} = 9a? + 2ab + OB’. 


AFFECTED QUADRATIC EQUATIONS. 


being a perfect square. 


This third term is the square of half the coefficient of x. 
Every affected quadratic may be made to assume the form 
x? + 2bx =e, by dividing the equation through by the co- 


efficient of 2”, 


QUADRATIC EQUATIONS. 253 


To solve such an equation : 


The first step is to add to both members the square of 
half the coefficient of x. This is called completing the square. 

The second step is to extract the square root of each mem- 
ber of the resulting equation. 

The third step is to reduce the aoe resulting simple 
equations. 


Ww 


(1) Solve the equation 2? — 8x = 20. 


We have x? — 8a = 20. 

Complete the square, x?— 8a +16 = 36. 

Extract the square root, xe—-4=+6. 

Reduce, e=4+6=10, 
or w=4—6=— 2, 


The roots are 10 and — 2; 


Verify by putting these numbers for w in the given equation 


2 e210, e=— 2, 
0? — 8(10) = 20, (— 2)? —8(—2) = 20, 
100— 80 = 20, 4 +16 = 20, 
(2) Solve the equation ets Sweat 
—]l a+ 9 | 
Free from fractions, (x + 1)(@ + 9) =(x—1)(42 —3). 


Simplify, 32? —17e#=6. 
We can reduce the equation to the form a?— 2 bz by dividing by 3. 
Divide by 3, a — lle = 2, 


Half the coefficient of x is } of — 17 = — 1%, and the square of — 4,7 
is 282. Add the square of — 1 to both sides, and we have 


Ard ete er ale Diet 289 





3 6 36 
eee 
2. 17 See Wes 
or x yor (2) 36 
17 19 


Extract the root, x — eins + 


wa, 


254 


Reduce, 


or 


SCHOOL ALGEBRA. 


Wii WALA 
rae ina 
17 19 36 
ig dee SOS gt 
TRE NRE Tg 


The roots are 6 and — 7 


Verify by putting these numbers for z in the original equation : 


x=6 
Baa aay 
Oe LG td 
7_ 21 
ee eS aanh's Cy 
Solve: 
1. 2 e220 = 8. 5. 
Sit the Hpk 
Sree == 1 a7, 
4,v7+42=5. 8. 





6. 








L 
4 
2) 
1 4 
ieee See 
133, eee 
iglhs aoe 
2 Sonu ds: 
4 26 
Exercise 93. 
2+ oa = 14. 9. # +227 =40 
x* — 8a = 28. 10. 82 —427=4 
20° -+ 2= 15, 11. 697 a a 
5 +- 8.22, 12. 62° —xz=2 


io a ea 


LA tLe) et) 


15. oe aot. La 
= ms gs ee 
4 6 


QUADRATIC EQUATIONS. 250 


bx+5 , 24—5 Pickens ede 
































ay f Os 22. . 

x+4 x—2 x—1 22 3 

18. x—6 TED es AS Be SOLAS 
zx—2 224+1 xa—-2 «+4 

i al a Bcc oy A 
BO Ee, 1-2 2 3 2x4—1 

DOr we 4#+1_13 OB. ate Lotion ae 18, 
zt+l1 x 6 g+2 “£+1.. 6 

21.52 —42—= 1. 26. T2?—-—82=—1. 


ANOTHER METHOD OF COMPLETING THE SQUARE. 


281. When the coefficient of 2? is not unity, we may 
proceed as in the preceding section, or we may complete 
the square by another method. 

Since (ax + 6)? = a’x’? + 2abs + 0’, it is evident that the 
expression a’x + 2abx lacks only the third term, b’, of being 
a complete square. 

It will be seen that this third term is the square of the 
quotient obtained from dividing the second term by twice the 
square root of the first term. 


282, Every affected quadratic may be made to assume 
the form of aa? + 2abxr =e. 


To solve such an equation : 


The first step is to complete the square; that is, to add to 
each side the square of the quotient obtained from dividing 
the second term by twice the square root of the first term. 

The second step is to extract the SEN root of each side 
of the resulting equation. 

The third and last step is to reduce the two resulting 
simple equations, 


256 SCHOOL ALGEBRA. 


(1) Solve the equation 1627+ 52—38=72?—2 + 46. 


We have 162° +5¢—3=72? —2 + 45. 
Simplify, 9a? + 6a = 48. 


To complete the square, take the square root of 927. This is 32, 
and twice 3a” is 6x. Divide the second term by 62, and the quotient 
is 1. Square 1 and add it to both sides. 

| 9a? +6n4+1= 49. 
Extract the square root, 32 +1=+7. 
Reduce, 3a — 1 +7 or 1 7, 
3a =6 or —8. 
*. w= 2 or — 22, 
Verify by substituting 2 for x in the equation: 
1627 +52—3 = 7a? — w + 45, 
16 (2)? + 5(2) ~ 3 = 7(2)? — (2) + 45, 
64 + 10—3 = 28 —2 +45, 
Thedh. 
Verify by substituting — 22 for x in the equation: 
1627+ 5¢%—3=72?—2 + 45, 


165) ee Mee een AGS 


1024 40 *, 448 8 
440 Te £a8h 8 Ae 
ae RNC x. 

1024 — 120 — 27 = 448 + 24 + 406, 

877 = 877° 


(2) Solve the equation 382?— 42 = 82. 


Since the exact root of 3, the coefficient of x2, cannot be found, it 
is necessary to multiply or divide each term of the equation by 3 to 
make the coefficient of a a square number. 


Multiply by 3, 9a? — 124 = 96. 
Complete the square, | 
9a? 124+4+4=100. 
Extract the square root, 37—2= + 10. 
Reduce, 32 2b 10 ore 10: 
3a = 12 or — 8. 
*. w= 4 or — 22, 


QUADRATIC EQUATIONS. 257 





a 4x 320 
Or, divide by 3, 5 (Ee gens ted 
Yr, Glvide by x 3 3 i 
4x ,4 32 4 _.100 
Complete th nl es 
SUES SS ARENE) up role mn 9 
Extract the square root, «— ; =+ rs 
2+ 10 
‘ _ t= ’ 
3 e 
= 4 or — 22. 
Verify by substituting 4 for x in the original equation: 
48 —16 = 32, 
32 = 32. 


_ Verify by substituting — 22 for x in the original equation: 
211 — (— 102) = 32, 
32 = 32. 
(3) Solve the equation — 32?+ 52 = — 2. 


Since the even root of a negative number is impossible, it is neces- 
sary to change the sign of each term. The resulting equation is 





3.0? — 5a = 2. 
Multiply by 3, 9a?— 152 =6. 
Complete the square, 
Ont = 15% 420 re” 
4 4 
Extract the square root, coe =+ 9 
Reduce, 32= ° 5 if 
32=6 or —1. 
. v2 = 2 or —= 
nat pe ee 
Or, divide b 2 —— = =. 
r, divide by 3, x anaes 
Complete the square, © 
| » ba 25 49 
g?— =~ 4 = —. 
3 36 36 


258 SCHOOL ALGEBRA.. 


Extract the square root, 2#— : =+ Z 
547 
L = —-, 
6 
= 2 or — 1 
3 


284, If the equation 32?— 5x = 2 be multiplied by four 
tumes the coefficient of x, fractions will be avoided. 


We have 36 2? — 60a = 24. 
Complete the square, 
36 a? — 60a + 25 = 49. 
Extract the square root, 6a —5=+ 7. 
62=54 7. 
6a = 12 or —2. 
p= 2Qor— 2 
3 
It will be observed that the number added to complete the square 
by this last method is the square of the coefficient of x in the original 
equation 32?—5a = 2. 
Nors. If the coefficient of x is an even number, we may multiply 
by the coefficient of «?, and add to each member the square of half the 
coefficient of # in the given equation. 


(1) Solve the equation 42?— 23% = — 30. 
o 


Multiply by four times the coefficient of a, and add to each side 
the square of the coefficient of 2, 


64 x? — () + (23)? = 529 — 480 = 49. 
Extract the square root, 8¢—23= + 7. 


Reduce, 8a = 2347; 
82 = 30 or 16. 
* @ == Stor a, 


Nore. If a trinomial is a perfect square, its root is found by taking 
the roots of the first and third terms and connecting them by the sign 
of the middle term. It is not necessary, therefore, in completing the 
square, to write the middle term, but its place may be indicated as 
in this example. 


QUADRATIC EQUATIONS. 259 


(2) Solve the equation 722? — 30% = — 7, 


Since 72 = 28 x 3?, if the equation is multiplied by 2, the coeffi- 
cient of x? in the resulting equation, 1442? — 60a = — 14, will be a 
square number, and the term required to complete the square will be 


(3:) . ale = Hence, if the original equation is multiplied by 


4 x 2, the coefficient of x? in the result will be a square number, and 
fractions will be avoided in the work. We shall then have 


576 x? — 240” = — 56. 
“57602 —() +25 =—31. 
Extract the root, Me sb = + V— Sl. 
a = (5 + V—831). 


Exercise 94. 


Solve: 
1. 3x'— 24= 8. 14. 30° +22 = 25 
2 OG 21. au 
$8OL Te, 
Ae Syed ser 1552 A da+1. 
4. 2a? 5x= 7. Tope ee ei 48D) 
5. 827°+7 6 oe.® j | 
6 aS he == Gi 
é a ee eee wa 
6. 52a? — 12 = 24. 3 9 
7. 827+32= 26. 18. OG oa E 
16 
8.0 (2-7 = 150, 20 
19. 8a°+ §2=—. 
9. 6a? +52=14. ; 3 
10. (2°—24= 4. 20. et aioe 
11. 82'+72=5]1. 21. fe 8 _ 20 
v 2 : ‘a Pas 3 rl Wars 
12 Oa 715, Oe Sa aaG 
Digi os ahs tare Ca. 
13. ll2?—102= 24. BM an 3 


260 


27. 


28. 


29. 


23. 


24. 








26. 2(52? — 8x —6) —4i(2’?— 3) = 2e+1. 
2 5 2 3 Za 
a 30. ro cele . 
pes Pe x—l p= eee 
Dene Nene 31. CTE oe ee 
x—-l 224+1 838 x—4 a-—2 
7 4 Q2—38 , 5—8e 
ee en =1. 
ee Oe Pees x+2 
ll—382 , 2(7—42) 
33.0 3 el 
l—2z a 1— 22 
nat epl ,la#_ ia 
w—4  -4+2% d(¢=—2) 
2a+7 , 84-2 
35. ==), 
ars z+1 
36 ADH DB fie iia are heed 
228-1) 2@ 4282 
2a—1 3 x—2 
37. — —__ => —___ 5, 
i 3 x—8 aaa 
38. B22 ee | 
2e—1 2a4+1 427-1 
—5 ,a2-—8 80 1 
30 a = = 
raga = 8 ae 20 28 
PTY. 2¢4+1 c+l1 45 ne 


SCHOOL ALGEBRA. 


(w + 2) (2% + 1) + (@— 1) (8a 4 2) = 57. 
382(2% + 5) — (x + 3)(82—1)=1. 


; SE MEE Lg, 
























































7—2 cee ~ 49 — 2 


QUADRATIC EQUATIONS. 261 


LITERAL QUADRATICS. 
285, Examples. 


(1) Solve the equation az?+ br+c=0. 


Transpose ¢, ax? + bx =—c. 
Multiply the equation by 4a and add the square of 5, 
4 atx? +() + 0? = 0? — 4a. 
Extract the root, 2ax+b=+ Vb? — 4ac. 
_=b+ VP—4ac 
2a 


(2) Solve the equation (a?+ 1)x=az’?+a. 
Transpose ax? and change the sign, 
ax? — (a? + 1)a=—a. 
Multiply by 4a, and complete the square, 
4atx?—()+ (a? +1P%=—4e@+at+ 2a? +1 


=at— 2a? +1. 
Extract the root, 2au—(a?+ 1) = + (a? — 1). 
Reduce, 2axz = (a? + 1) + (a? — 1), 
= 2a? or 2. 
: 1 
. &= a or — 
a 


(3) Solve the equation adx — aca’ = bex — bd. 


Transpose bcx and change the signs, 

acx? + bex — ada = bd. 
Express the left member in two terms, 

acu? + (be — ad) x = bd. 
Multiply by 4ac, and complete the square, 
4a%ctx? + () + (be — ad)? = bc? + 2abed + a?d?. 

Extract the root, 

2ace + (be — ad) = + (be + ad). 


Reduce, 2acx = — (be — ad) + (be + ad) 
= 2ad or — 2be. 
d b 
_£=  or—-- 


C a 


262 SCHOOL ALGEBRA. 


(4) Solve the equation pa? — px + ga? + qr = are 
Express the left member in two terms, | 
at Bs 
Pag 
Multiply by four times the coefficient of 2?, 

4(p + ga! — 4(p? — q?) a = £pg. 
Complete the square, 

4(p + ga? —() + (p —g =p? + 2pg + 9. 
Extract the root, 
2(p + q)a—(p—q)= + (p + 9). 
Reduce, 2(p + q)e=(p—9)+(p +9), 
= 2p or — 24. 


pe ee eee 
P+q Pec’ 


(p+ 9)” —(p—g)x= 


Note. The left-hand member of the equation when simplified 
must be expressed in two terms, simple or compound, one term con- 
taining x, and the other term containing a. 


Exercise 95. 


Solve: 
le 2 2av OG; 9. 272’? + ax—1=0. 
eat A ah ahs 10. 120727—5b2=8. 
3. 2 ++8b2 =90D". 11. 22 4.11 a(v—8a). 
4. 2°+3b72 =1086’. . ea ta 
5. 27+ 5axz = 14a’. + aie Ae 
6. 8247+ 4er = 4c’. Tg 5? ee 
a 4a 
: Een OA ha 
7 5 ax x a 7 Baa’ 2n_ 18. 
8. 627—ar—av=0. Pie 3 a 
2 
15. oma aie a Mee Bo 


3 4 


16. 


17. 


18: 


19. 


20. —— 


21. 


37. 
38. 
39. 
40. 


22 —2e2-4+G=207 22. 

w(a@—%),%_» oe 

ata 3 

u(d%—a)_ a 24, 

4z+a lye 

aes EE ssh 25. 
mn 

Poe Gee Alone 3 26: 

b—a «+a 

a+ 20 | OO wey 

36—x2 «+20 


QUADRATIC EQUATIONS. 





2+ (a—b)x=ab. 


2atea ie Pao 
2a—x% a+2x 88 
22—30 , 3@+2a 10) 








x+4a 4z—a 


ve to ae be 
i ee ee 





29. 
30. 
31. 
32. 
33. 
34. 


35. 


36. 


9a°— 8(a+2b) 2+ 2ab=0. 
(2a+1)7?+38ae4+a—av=0. 

(1 — a’) 2? — 2(1+a’)4+1—a’=0. 
(a+ 5)? — (a? — 0?) 2 = ab. 

vet Qbear ta=e'2? +2e7r+ 6". 
(a+ b6)2 — (2a+6)rx+a=0. 


Qe—8b 8a 


xz—26b 





x—b 


SD eA (ae by 


(3a? + 0?) (2? —x+1)= (#7 +30’) (2+ 2+41). 
“—(a+b)e+ac+be—e?=0. 

x — pa=(p+q+r)(q+7). 

(a? + ab) (x? —1) — (av + 0’) 2+ (a+ 6) (a—26)=0. 


263 


264 SCHOOL ALGEBRA. 


SOLUTIONS BY FACTORING. 


286. A quadratic which has been reduced to its simplest 
form, and has all its terms written on one side, may often 
have that side resolved by znspection into factors. 

In this case, the roots are seen at once without com- 
pleting the square. 


(1) Solve x? + 72 —60=0. 


Since a + 7x — 60 = (x + 12) (#% — 5), 
the equation x? + Tx —60=0 
may be written (x + 12)(#—5) =0. 


If either of the factors x +12 or «—5 is 0, the product of the two 
factors is 0, and the equation is satisfied. 


Hence, x+12=0, and z—5=0. 
*, e=—12, and «=5. 

(2) Solve 22° — 2? —6x%=0. 
The equation 20° — 2? -62=—0 
becomes x(2x? — 2 —6) =0, 


and is satished if a2==O0, or if 2227-x2-6=0. 


By solving 22? —- x—6=0, the two roots 2 and — : are found. 


Hence the equation has three roots, 0, 2, — . 


(3) Solve 2+ 2?--4z% —4=0. 
The equation a +a?-4r—-4=0 
becomes x (~4 +1)—4(x + 1)=0, 
(x? —4)(@ + 1) =0. 
Hence the roots of the equation are — 1, 2, — 2. 


(4) Solve 2 —22?—11lz+12=0. 
By trial we find that 1 satisfies the equation, and is therefore a 
root (¢ 89). 
Divide by 2 —1; the given equation may be written 
(x -- 1)(a? — 2 — 12) = 0, 
and is satisfied if s —1=0, or if r? -x—12=0, 
The roots are found to be 1, 4, — 3. 


QUADRATIC EQUATIONS. 265 


(6) Solve the equation x(2? — 9) =a(a’ — 9). 
If we put a for a, the equation is satisfied; therefore a is a 
root (2 89). 
Transpose all the terms to the left-hand member and divide by 
x — a. 
The given equation may be written 
(a — a) (x? + ax + a? — 9) =0, 
and is satisfied if s—a=0, 
or if 27+ ar+a?—9=0. 
The roots are found to be 
a, —a+ V36 — 3a? —a— V36 — 3a? 
2 2 


Exercise 96. 


Find all the roots of 


1. 2?—57+4=0. 5. a+ 2?—62=0. 
2. 627—5x—6=0. 6. 2 —8=0. 
3. 227—x—-—38=0. (eri So 0, 
4. 10¢’?+2—3=0. 8. 2 —16=0. 


9. («—1)(x—8)(27+52+46)=0. 

10. (22 —1)(# — 2) (82? —- 5x — 2) =0. 

ll. (@?+4— 2)(22?+32—5)=0. 

12. #+27—4(r4+1)=0. 

18. 32°+ 227 — (82+ 2)=0. 

14. 2#—27—1382+ 39=0. 
15. +84 3(27—4)=0. 17. 2a°—2Q2*—(2?—1)=0. 
16. «(@—1)—6(@—1)=0. 18. “-—32—2=0. 

19. 2a°+ 22’ + (a? — 57 —6)=0. 

20. “#—427*?—2’?+162-—12=0. 


266 SCHOOL ALGEBRA. 


SOLUTIONS BY A FORMULA. 


287. Every quadratic equation can be made to assume 
the form az’?+ ba+e=0 
Solving this equation (§ 285, Ex. 1), we obtain for its 
two roots 
Sait — b= NV bh age 


—A4Aace 4 ac. 
; 2a 


2a 


There are two roots, and but two roots, since there are 
two, and but two, square roots of the expression b?— 4ac. 

By this formula, the values of x in an equation of the 
form az’-+ bx +¢=0 may be written at once. 


Ex. Find the roots of the equation 8a7— 5x 4+2=0. 
Here a=3, b=—5, c=2. 


Putting these values for the letters in the above formulas, we have 


put V2 —24 4, 5— V2 24 
6 6 
=§ org 
=1 or 2 


Exercise 97. 


Solve by the above formulas: 


1. 22°+382=14. 7 52°—Te#=—2. 
2. 82-02 = 12. 8. 427-924 = 28. 

3. @—Te= 18. 9.4 Dat Te 12. 

4. Bat— 2 = 42, 10. 112? ~92=— 
5. 627—Tx=10:"' 11. 7a7?+- 5a = 88. 
6. 827—llx=—6. 12. (bine eG) 


QUADRATIC EQUATIONS. \ AG | 


EQUATIONS IN THE QUADRATIC Form, 
eX 


‘Ee Yh 
‘~o 


a 


288. An equation is in the quadratic form if it ¢ 
but two powers of the unknown number, and the exponent 
of one power is exactly twice that of the other power. 






289. Equations not of the second degree, but of: the 
quadratic form, may be solved by completing the square. 


(1) Solve: 82° + 632° = 8, 
This equation is in the quadratic form if we regard a? as the un- 
known number. 
We have 82% + 6323 = 8. 
Multiply by 32 and complete the square, 
256 2° + () + (63)? = 4225. 
Extract the square root, 1623 + 63 = + 65. 
Hence, oF me : or — 8. 


Extracting the cube root, two values of are} and —2. There 
are four other values of « which we do uot find at present. 


(2) Solve: Vat — 8/28 = 40. 
Using fractional exponents, we have 
a? — 3a% = 40, 


This equation is in the quadratic form if we regard 2 as the un- 
known number. 


Complete the square, 42?—1224+9 = 169. 


Extract the root, 204 —3-= + 13, 
~. 20% = 16 or — 10, 
at = 8 or —5, 
¢ = 16 or —5 V5. 


There are other values of « which we do not find at present. 


268 SCHOOL ALGEBRA. 


(3) Solve completely the equation z*= 1. 


We have x —1=0., 
Factoring, (#—1)(#+a4+4+1)=0. 
Therefore, either x—1=0 
or et+er+1=0. 
ZS) Bs a? +a+1=0, 
a= 1, ; wa ere Moa 
Solving, «= ar ts oa 
The three values, 1, alt Vans, FTE ein aia best are the three cube 
2 2 
roots of 1. 
(4) Solve: (22 — 3)? — (22 — 3) = 6. 
Put y for 2% —3, and therefore y? for (2” — 3)”. 
We have y?—y =6. 
Solving, y =3 or —2. 
Putting now 2x—8 for y, 
oe Ses Ine 
v= 3. v=. 


Exercise 98. 


1. af —5att4=0. 6. 10a#—21=2". 

2. 2'— 1842+ 36 =0. 7. V2'+8V2=18. 

8. at—212?=100. 8. BV 2 —2V «= — 20. 

4. 49° — 32° = 27. 9. 52°"+ 32" = 68. 

5. Qat+ 5a? = 218. 10. (82+3)*4+ (8248) =30. 
11. 2(@?@—241)—V2?—-2+1=1. 

12. 2&—92+8=0. . 14. (e@+1)4+vV2+1=6. 

13. fhe tat 15: 2° 13827 = - 36) 


16. 2¢7?+47+948V227+42+9=40. 


122° — 1llz?+ 10%— V8 ] 
17 ae a eS ee eee 
82°— Tx+6 be 2 


QUADRATIC EQUATIONS. 269 


RapDiIcaAL EQUATIONS. 


290. If an equation involves radical expressions, we first 
clear of radicals as follows: 


Solve Vz +4+V22+6=Vi72-+ 14. 

Square both sides, 

; e2+4+2V(e2+4)(22+6)+204+6=7e +14. 
Transpose and combine, 2V(« + 4) (224+ 6) =40 +44. 


Divide by 2 and square, (a + 4)(2” + 6) = (2x 4 2)%, 
Multiply out and reduce, x? — 3a” = 10. 
Hence, x= 5 or — 2, 


Of these two values, only 5 will satisfy the original equation, 
The value — 2 will satisfy the equation 


Ve+4~—V2024+6=Vi70414. 
In fact, squaring both members of the original equation is equiva- 
lent to transposing 1/72 +14 to the left member, and then multiplying 


by the rationalizing factor Vz +4 + V22+6+ W72«+4+ 14, so that 
the equation stands 


(V2+44+ V204+6— Via + 14)(V24+44+ V204+6 4+ V72+14) =0, 
which reduces to V(x + 4)(2a + 6) —(2a + 2) =0. 
Transposing and squaring again is equivalent to multiplying by 
ei an Cee eee VERE OV ibe 1a) 
Multiplying out and reducing, we have 
x? —3x2—10=0. 
Therefore, the equation 2? —-3a—10 = 0 is really obtained from 
(Sew ay Wie te 7 8) 
x (V2+44+ V224+64+ V72 +14) 
x (Ve +4—V2024+6— Via +14) 
x (Vat 4—V20+64+ Vix +14) =0. 
This equation is satisfied by any value that will satisfy any one 
of the four factors of its left member. The first factor is satisfied 


by 5, and the last factor by — 2, while no values can be found to 
satisfy the second or third factor. 





O70 : SCHOOL ALGEBRA. 


Hence, if a radical equation of this form is proposed for solution, 
if there is a value of # that will satisfy the particular equation given, 
that value must be retained, and any value that does not satisfy the 
equation given must be rejected. (See Wentworth, McLellan and 
Glashan’s Algebraic Analysis, pp. 278-281.) 


291. Some radical equations may be solved as follows: 
Solve 72? —524+-8V727?—5a#+1=—8. 
Add 1 to both sides, 


Tot — 5a +14 8V7e?—524+1=—7. 
Put V72?—52+1=y; the equation becomes 


y? + 8y=— 7. 
Hence, y =—lor—7, 
<= loor 49. 


We now have 72?—5a2+1=1, or 72?~—524+1= 49, 
Solving these, we find for the values of 2, 
peo iele. 
7 7 

These values all satisfy the given equation when we take the 
negative value of the square root of the expression 7a7—5a2+1; 
they are in fact the four roots of the biquadratic obtained by clear- 
ing the given equation of radicals. 


Exercise 99. 
Solve: 


1 V92+404+ 2V2et+7= Vz. 
as VasG Vas2=V 6. 


824+ V4e—H _o 
82—WV4ae— 2° 


4, Vz—8—V2r—14=~V4a — 155. 
5. Vet4—Ve=Vat8: 


22. 


23. 


QUADRATIC EQUATIONS. 211 


oY pet ED ONE 
Q2+Ve 404+Vz 
Vee OOS on/ ee n/8 5 oS 0: 


fae ee any rl e (), 





. Ve—24+ V2e+3— V4e4+1=0. 

NA Sa A308 a8 = 0, 

. BV 4174+ Ve +14 2V5e+ 41 =0. 
. 22 —-V22—-1l=2+42. 

BEY CED N/a LNs 0, 


1 1 





ee ee ee 
BN yee lee aN) aL 
eee Seals eee 
V3%+ 13 
.2—-Ve—22'—2=0. 
Ce NE ss anh Vers 
ae SY een 


ooo 87 pV Oe + or 49 — 8. 
Rea ha N/a el == OD. 

ee So btn) Oat Bot 
Oe Oe 8 a tO 


82°— 424+ V32?—42—6=18. 


82°7°— 74+ 8V32? — 1624+ 21 = 162. 


22 SCHOOL ALGEBRA. 


292. Problems involving Quadratics. Problems which in- 
volve quadratic equations apparently have two solutions, 
since a quadratic equation has two roots. When both roots 
of the quadratic equation are positive integers, they will 
give two actual solutions of the problem. 

Fractional and negative roots will in some problems give 
admissible solutions; in other problems Bory. will not give 
admissible solutions. 

No difficulty will be found in selecting the result which 
belongs to the particular problem we are solving. Some- 
times, by a change in the statement of the problem, we may 
form a new problem which corresponds to the result that 
was inapplicable to the original problem. 

Imaginary roots indicate that the problem is impossible. 

Here as in simple equations x stands for an unknown 
number. 


(1) The sum of the squares of two consecutive numbers 
is 481. Find the numbers. 


Let x = one number, 
and x + 1 = the other. 
Then + (« + 1)? = 481, 
or aie ane 


The solution of which gives = 15 or —16. 

The positive 15 gives for the numbers, 15 and 16. 

The negative root — 16 is inapplicable to the problem, as consecu- 
tive numbers are understood to be integers which follow one another 
in the common scale, 1, 2, 3, 4..... 


(2) A pedler bought a number of knives for $2.40. 
Had he bought 4 more for the same money, he would have 
paid 3 cents less for each. How many knives did he buy, 
and what did he pay for each? 


Let x = number of knives he bought. 


Than: a = number of cents he paid for each. 


QUADRATIC EQUATIONS. ON Ga 








But if x + 4= number of knives he bought, 
240 _ number of cents he paid for each, 
G+ 
240 _ _240 _ the difference in price. 
af PS ry 
But 3 = the difference in price. 
240240 _ 
ee ed 
Solving, x= 16 or — 20. 


He bought 16 knives, therefore, and paid 249 or 15 
cents for each. 
If the problem is changed so as to read: A pedler bought 
a number of knives for $2.40, and if he had bought 4 /ess 
for the same money, he would have paid 3 cents more for 
each, the equation will be 
240 240 
| ae Se 
Solving, x= 20 or — 16. 
This second problem is therefore the one which the neg- 
ative answer of the first problem suggests. 


(3) What is the price of eggs per dozen when 2 more in 
a shilling’s worth lowers the price 1 penny per dozen? 





Let a = number of eggs for a shilling. 
Then == cost of | egg in shillings, 
and 12 cost of Tidozeniin shillinpa: 
2 
But if x + 2=number of eggs for a shilling, 
12 cost of 1 dozen in shillings. 
© + 2 
tomes 


= i (1 penny being +; of a shilling). 


The solution of which gives «= 16, or — 18. 
And, if 16 eggs cost a shilling, 1 dozen will cost 9 pence. 


Therefore, the price of the eggs is 9 pence per dozen. 


274 SCHOOL ALGEBRA. 


If the problem is changed so as to read: What is the 
price of eggs per dozen when two Jess in a shilling’s worth 
rawes the price 1 penny per dozen? the equation will be 


12 the of a 


—_— — _ 





The solution of which gives «= 18, or — 16. 
Hence, the number 18, which had a negative sign and was inappli- 
cable in the original problem, is here the true result. 


Exercise 100. 


1. The sum of the squares of two consecutive integers is 
761. Find the numbers. . 


2. The sum of the squares of two consecutive numbers ex- 
ceeds the product of the numbers by 13. Find the numbers. 


3. The square of the sum of two consecutive even num- 
bers exceeds the sum of their squares by 336. Wind the 
numbers. 

4. Twice the product of two consecutive numbers ex- 
ceeds the sum of the numbers by 49. Find the numbers. 


5. The sum of the squares of three consecutive numbers 


is 110. Find the numbers. 


6. The difference of the cubes of two successive odd 
numbers is 602. Find the numbers. 


7. The length of a rectangular field exceeds its breadth 
by 2 rods. If the length and breadth of the field were 
each increased by 4 rods, the area would be 80 square rods. 
Find the dimensions of the field. 


8. The area of a square may be doubled by increasing 
its length by 10 feet and its breadth by 38 feet. Determine 
its side. 


QUADRATIC EQUATIONS. 275 


9. A grass plot 12 yards long and 9 yards wide has a 
path around it. The area of the path is 2 of the area of the 
plot. Find the width of the path. 


10. The perimeter of a rectangular field is 60 rods. Its 
area 1s 200 square rods. Find its dimensions. 


11. The length of a rectangular plot is 10 rods more 
than twice its width, and the length of a diagonal of the 
plot is 25 rods. What are the dimensions of the field? 


12. The denominator of a certain fraction exceeds the 
numerator by 8. If both numerator and denominator be 
increased by 4, the fraction will be increased by 3. Deter- 
mine the fraction. 


13. The numerator of a fraction exceeds twice the de- 
nominator by 1. If the numerator be decreased by 38, and 
the denominator increased by 3, the resulting fraction will 
be the reciprocal of the given fraction. Find the fraction. 


14. A farmer sold a number of sheep for $120. If he 
had sold 5 less for the same money, he would have received 
$2 more per sheep. How much did he receive per sheep? 

State the problem to which the negative solution applies. 


15. A merchant sold a certain number of yards of silk 
for $40.50. If he had sold 9 yards more for the same 
money, he would have received 75 cents less per yard. 
How many yards did he sell? 


16. A man bought a number of geese for $27. He sold 
all but 2 for $25, thus gaining 25 cents on each goose sold. 
How many geese did he buy? 


17. A man agrees to do a piece of work for $48. It 
takes him 4 days longer than he expected, and he finds 
that he has earned $1 less per day than he expected. In 
how many days did he expect to do the work? 


276 SCHOOL ALGEBRA. 


18. Find the price of eggs per dozen when 10 more in 
one dollar’s worth lowers the price 4 cents a dozen. 


19. A man sold a horse for $171, and gained as many 
per cent on the sale as the horse cost dollars. How much 
did the horse cost? 


20. A drover bought a certain number of sheep for $160. 
He kept 4, and sold the remainder for $10.60 per head, and 
made on his investment # as many per cent as he paid dollars 
for each sheep bought. How many sheep did he buy? 


21. Two pipes running together can fill a cistern in 52 
hours. The larger pipe will fill the cistern in 4 hours less 
time than the smaller. How long will it take each pipe 
running alone to fill the cistern ? 


22. A and Bcan do a piece of work together in 18 days, 
and it takes B 15 days longer to do it alone than it does A. 
In how many days can each do it alone? 


23. A boat’s crew row 4 miles down a river and back 
again in 1 hour and 30 minutes. Their rate in still water 
is 2 miles an hour faster than twice the rate of the current. 
Find the rate of the crew and the rate of the current. 


24. A number is formed by two digits. The units’ digit 
is 2 more than the square of half the tens’ digit, and if 18 
be added to the number, the order of the digits will be 
reversed. Find the number. 


25. A circular grass plot is surrounded by a path of a 
uniform width of 3 feet. The area of the path is ¢ the area 
of the plot. Find the radius of the plot. 


26. Ifa carriage wheel 11 feet round took 4 of a second 
less to revolve, the rate of the carriage would be 5 miles 
more per hour. At what rate is the carriage travelliny ? 


CHAPTER XIX. 


SIMULTANEOUS QUADRATIC EQUATIONS. 


293, Quadratic equations involving two unknown num- 
bers require different methods for their solution, according 
to the form of the equations. 


Case I. 


294, When from one of the equations the value of one of the 
unknown numbers can be found in terms of the other, and this 
value substituted in the other equation. 


Ex. Solve: 32° — 2Qay=5 \ (1) 
r—y=2 (2) 
Transpose 2 in (2), y=x— 2, 


In (1) put x — 2 for y, 
3a? — 2a(e —2)=5, 


The solution of which gives e=1, or x= —5. 
If x=1, 
y=1—2=-1; 
and if e=—5, 


We have therefore the pairs of values, 


2=1 e=—5 
a6 
y=—1 Fi ei 7 
The original equations are both satisfied by either pair of values. 
But the values «=1, y=—7, will not satisfy the equations; nor will 


the values # =—5, y=—1. 


The student must be careful to join to each value of x 
the corresponding value of v. 


278 SCHOOL ALGEBRA. 


Case II. 


295. When the left side of each of the two equations is 
homogeneous and of the second degree. 





Solve: eT: i (1) 
y—v2=16 (2) 
Let y = va, and substitute vx for y in both equations. 
From (1), 2v°a? — 40a? + 3a? = 17. 
Bede ti Ll lig! 
2v?—40+4+3 
From (2), vy? — a = 16. 
= 16 
v— 1 
Equate the values of 2”, oe a POs 





Qv%— 4943 v1 
320? — 640 + 48 = 1772-17, 
150? — 64v = — 65, 
225 v? — 960 v = — 975, 
225 v? ~ () + (32)? = 49, 





15v— 32 = + 7. 

5 13 

v= or —- 

3 5 
3 5 
y= v= 22 y= ve = 258 

Substitute in (2), Substitute in (2), 

250" 2-16 160 = 16 

9 ; 25 
a? = 9, pS, 

=+3 

ne + 8, ers 
erie otuali 
jong 


SIMULTANEOUS QUADRATIC EQUATIONS. 279 


Case ITI. 


296. When the two equations are symmetrical with respect to 


x and y; that is, when x and y are similarly involved. 
Thus, the expressions » 


204-38 27y'?+2y', 2ay—8xe—8y+4+1, 2t-8ay—8ay+y', 


are symmetrical expressions. In this case the general rule 
is to combine the equations in such a manner as to remove 
the highest powers of # and y. 


Solve: at + y' = 337 \ (1) 
ee pee a) (2) 


To remove x‘ and 74, raise (2) to the fourth power, 


at + 4a8y + 6a?y? + 4ay2+ y* = 2401 
Add (1), oe + yt= 337 
Qat + 4a8y + 6a?y? + 4ay? + 2y* = 2738 


Divide by 2, at + 2a°y + 3a%y? + 2ay>+ y* = 1369. 

Extract the square root, a+ ay +y? = + 37. (3) 
Subtract (3) from (2)?, ay = 12 or 86. 

We now have to solve the two pairs of equations, 


ety= 7). @+y= ry. 


ay=125' ay = 86 
From the first, ~ ee St my OA 
y=3 y=4 
From the second, igs i 
Gee = 295 
—— 


2 


280 SCHOOL ALGEBRA. 


297, The preceding cases are general methods for the 
solution of equations which belong to the kinds referred to; 
often, however, in the solution of these and other kinds of 
simultaneous equations involving quadratics, a little inge- 
nuity will suggest some step by which the roots -may be 
found more easily than by the general method. 


(1) Solve: ae aa (1) 
xy = 300 (2) 
Square (1), a + 2ay + y? = 1600. (3) 
Multiply (2) by 4, 4 xy = 1200. (4) 
Subtract (4) from (3), 
x? — 2ay + y? = 400. (5) 
Extract root of each side, «—y = + 20. (6) 
From (1) and (6), 2=30)>  ¢= 10 


jolts S00n 


1 ee *) 
2) Solve: = + ==— (1) 
(2) t+i=3 
Leh oe aL . 
=e aa SS ees 9 
es cp y® 400 @) 
Leute wile eer 
8 f Sats typsecy ge hes 2 ph 194% 3 
se Ga way" y® 400 ©) 
Subtract (2) from (38), ae a (4) 
Subtract (4) from (2), 
me Spin ge 
lena ime ke 
Extract the root, Sy neki (5) 
x 0 


From (1) and (5), 


SIMULTANEOUS QUADRATIC EQUATIONS, 281 


(3) Solve: z—y= 4 \ (1) 

e+y = 40 (2) 
Square (1), x? —2ey + y? = 16. (3) 
Subtract (2) from (3), — 2ay =— 24, (4) 


Subtract (4) from (2), 
x + 2ey + y? = 64. 


Extract the root, e+y=+8. (5) 
From (1) and (5), e= 6) ) ¢=— 2). 
y=2 y=—6 
(4) Solve: ey en Ol \ (1) 
Coty (2) 
Divide (1) by (2),  # —ay+y?=13, (3) 
Square (2), x? + 2ey + y? = 49. (4) 
Subtract (3) from (4), say = 36. 
Divide by —3, —xy=—12. (5) 
Add (5) and (8), a? ~ 2ay+y?=1. 
Extract the root, e—y=+1. (6) 
From (2) and (6), aaa cel eal 
y=3)’ y= 
(5) Solve: oa == 18 ay } (1) 
aie ae (2) 
Divide (1) by (2), a? —ay + y2# =22U. (3) 
Square (2), x + Qey + y? = 144. (4) 
Subtract (4) from (3),  —3ay= au ~ 144, 
» which gives ay = 32, 
We now have, e+ty=12). 
xy = 32 


Solving, we find, n= 8 \ tae a ae 
y=4 y=8 


282 


10. 


it. 


1%. 


14. 


15. 


16. 


2 -—-y =] 


SCHOOL ALGEBRA. 


Exercise 101. 


: ie, 3. eae 5 PAG Sa 
zy lG zy =— 8 x+y? = 80 
alae 4. ae oat 6. «ty = 3) 
ry = 27 xy=11 a? + 4? = 29 § 
ZY = ze 18. et eae 
xv? +-y? = 45 s2—y=3 

oh ik he a ee 
vt y=10 rye 
. 8a —y =12 bee 
a —y*?=16 ey 
y=32+1 i 20. 1,1_5 
2 + xy = 33 La he eo 8 
Gs aston 5+4=5 
847-477 = 8 gti 
eas, 21. regu fs 
cy 6 cee 
22 —8y =2 ay = 6 
v—22y=—i7 
22. elas 
23 = “1 Paes, 
32° — 4ay = 82 1,1_6 
x y° 
a? — xy + y= 21) 
=9 
ge a eae vee 
22-34 = 7 xy =16 
. ova 24. a+43 = 85 


SIMULTANEOUS QUADRATIC EQUATIONS. 288 


: eee eae) 37. 2 +ay+y’'=39 ; 
a2—y= 1 227+ 8ay+ 7? = 63 
eves 38. eee 
FY 20 ay + 2y* = 40 
2. pies 
° pi ewes 39, af eas 
aes ay +y=14 
. &-Y=? 
ae at 40. vtay+27=44 
Mery ea 4 
20°—8ayt2y’?=16 


Sy =126 


oe —ay ty = 21 41. #4+37=81 i 


ae af == 90 SPY Si ane 
Lp ey Hy mye: 42. 327+ Tay = 82 
Gh fn oO a? + 5ay + 9y’? = 279 
ais: |? 

43. g'+ty'=97) 

i Lill a+ty= 5) 
Sat ame a 
Aa 44. a eat 
Le a+y= 8 
fag a Nyy |x! 
y #16 Beer we: 
ze Metres) 46. REA 
xy + y?= 40 Bes pacetiat 

—e —ay = : 47. mitate «| 
sy- = Boa ioe tic 

. &+2ry= 24 48. Sea ee 
2ry +47? = 120 2a+y =zy—1 

ip Be ay tt 49. pga. st 
Tay+t9y = 50 z—-y=2 


284 SCHOOL ALGEBRA. 


BO. Hee, eee 61. Mee ae 
51. EL ce 
ee ced Bice 
3x’y + y* = 63 62 ela wale 
Bay Lge i Se Se a malo 
a+tay + y= 1281 rie 


53. xr— xy +y= Sf Ron a? = an by 
x + ay? + y* = 21 ; 


it Y toe 7) 

















ee ts 64. ede te 
Rane PM Pee xy =a? — 2 
vty? = 20 
4 4 
65. a2 , Pi +6 ] 
Bb. oe a aah Ys ee: cat! ab? 
Papa yD: 
zy == 1 
Sa +4y = 36 
56. PAGS LEN | 66. a ore 
sy +16=0 : 
ys ab 
57. r—y—3=0 I 4 (aby 
2 (2? — y’) = Bay 
67. 2a Piet | 
Ba) Gece ay — y= Qab— 2B 
ee is 
1 Le eat 68. ea eis) 
of aly vy = 6 


59. af ty! = 272 69. pa 
ey = 382y—4 xy =a? —4B° 


60. V@ty=2y a 70. Barbe er 
za+y =2ry—1 zx+y=a+b 


SIMULTANEOUS QUADRATIC EQUATIONS. 285 


Exercise 102. 


1. The area of a rectangle is 60 square feet, and its 
perimeter is 34 feet. Find the length and breadth of the 
rectangle. 


2. The area of a rectangle is 108 square feet. If the 
length and breadth of the rectangle are each increased by 
3 feet, the area will be 180 square feet. Find the length 
and breadth of the rectangle. 


3. If the length and breadth of a rectangular plot are 
each increased by 10 feet, the area will be increased by 400 
square feet. But if the length and breadth are each dimin- 
ished by 5 feet, the area will be 75 square feet. Find the 
length and breadth of the plot. 


4. The area of a rectangle is 168 square feet, and the 
length of its diagonal is 25 feet. Find the length and 
breadth of the rectangle. 


5. The diagonal of a rectangle is 25 inches. If the 
rectangle were 4 inches shorter and 8 inches wider, the 
diagonal would still be 25 inches. Find the area of 
the rectangle. 


6. A rectangular field, containing 180 square rods, is 
surrounded by a road 1 rod wide. ‘The area of the road 
is 58 square rods. Find the dimensions of the field. 


7. Two square gardens have a total surface of 2137 
square yards. A rectangular piece of land whose dimen- 
sions are respectively equal to the sides of the two squares, 
will have 1098 square yards less than the two gardens 
united. What are the sides of the two squares? 


8. The sum of two numbers is 22, and the difference 
of their squares is 44. Find the numbers. 


286 SCHOOL ALGEBRA. 


9. The: difference of two numbers is 6, and their 
product exceeds their sum by 39. Find the numbers. 


10. The sum of two numbers is equal to the difference 
of their squares, and the product of the numbers exceeds 
twice their sum by 2. Find the numbers. 


11. The sum of two numbers is 20, and the sum of their 
cubes is 2060. Find the numbers. 


12. The difference of two numbers is 5, and the differ- 
~ence of their cubes exceeds the difference of their squares 
by 1290. Find the numbers. 


13. A number is formed of two digits. The sum of the 
squares of the digits is 58. If twelve times the units’ 
digit be subtracted from the number, the order of the 
digits will be reversed. Find the number. 


14, A number is formed of three digits, the third digit 
being twice the sum of the other two. The first digit plus 
the product of the other two digits is 25. If 180 be added 
to the number, the order of the first and second digits will 
be reversed. Find the number. 


15. There are two numbers formed of the same two 
digits in reverse orders. The sum of the numbers is 33 
times the difference of the two digits, and the difference of 
the squares of the numbers is 4752. Find the numbers. 


16. The sum of the numerator and denominator of a cer- 
tain fraction is 5; and if the numerator and denominator 
be each increased by 38, the value of the fraction will be 
increased by 3. Find the fraction. 


17. The fore wheel of a carriage turns in a mile 1382 
times more than the hind wheel; but if the circumferences 
were each increased by 2 feet, it would turn only 88 times 
more. Find the circumference of each, 


CHAPTER XX. 
PROPERTIES OF QUADRATICS. 


298. Every affected quadratic can be reduced to the form 
ax’ + bx + ¢=0, of which the two roots are 


b , Vi'—Aac b Vbi—4ac ¢ 
Beae ee Yeoman Nee, eR OR 
ere nage i 2a 2a : 


CHARACTER OF THE Roots. 


299. As regards the character of the two roots, there are 
three cases to be distinguished. 


I, If b’—4ac is positive and not zero. In this case the 
roots are real and unequal. The roots are real, since the 
square root of a positive number can be found exactly or 
approximately. If 6*—4ac is a perfect square, the roots 
are rational; if 6?—4ace is not a perfect square, the roots 
are surds. 

The roots are unequal, since Vb? — 4ac is not zero. 


II. If b?— 4ac is zero. In this case the two roots are 


real and equal, since they both become on 
a 


IIT, If b’—4ac is negative. In this case the roots are 
wmaginary, since they both involve the square root of a 
negative number. 

The two imaginary roots of a quadratic cannot be equal, 
since 6?—4ae is not zero, They have, however, the same 


288 SCHOOL ALGEBRA. 


real part, ~2". and the same imaginary parts, but with 
a 


opposite signs; such expressions are called conjugate im- 
aginaries. The expression 6’— 4ac is called the discriminant 
of the expression az?-+ bx+e. 


300. The above cases may also be distinguished as follows: 


CasE I. &?—4ac>0, roots real and unequal. 
CasgE II. 6?—4ac=0, roots real and equal. 
CasE III. &?—4ac <0, roots imaginary. 


301. By calculating the value of 6’—4ae we can deter- 
mine the character of the roots of a given equation without 
solving the equation. 


(La — oa 60. 


Here a=1, b=—5d, ¢=6. 
b?—4ac = 25 — 24=1. 


The roots are real and unequal, and rational. 


(2) 827+ 72—-1=0. 


Here a=3, b=7, c=—1, 
b?}~4ac =49 4+ 12=—61. 


The roots are real and unequal, and are both surds. 


(3) 42?—122+9=0. 


Here a=4, b=—12, c=~—9. 
b?—4ac=144—144=0. 


The roots are real and equal. ‘ 
(4) 22—82t+4=—0. 


Here a=2, b=~—3, c=4. 
b?—4ac=9 — 32 = — 23. 


The roots are both imaginary. 





PROPERTIES OF QUADRATICS. 289 
(5) Find the values of m for which the following equa- 
tion has its two roots equal : 
2mzx? + (5m + 2)2+ (4m+1)=0. 
Here a=2m, b=5m+2, c=4m+1. 
If the roots are to be equal, we must have 
b? —4ac=0, or (5m + 2)?—8m(4m + 1) =0. 
This gives m= 2, or — = 
For these values of m.the equation becomes 


4¢74.122+9=0, and 427— 424+1=0, 


each of which has its roots equal. 


Exercise 103. 


Determine without solving the character of the roots of 
each of the following equations: 


¢’tbat+6=0. 6. 62°? —Ta—3=0. 
v’+2a2—15=0. 7. 527—dxe4—8=0. 
vtQer+3=0. nn 8. 2a?—x+5=0, 

8a + Ta#+2=0. 9. 62° +42—T7=0. 
927+ 6x4+1=0. 10. Bat + 80+" = 0, 


Determine the values of m for which the two roots of 
each of the following equations are equal: 


11. (m+)2?4+ (m—1l)ae+m4+1=0. 
12. (2m—3)2?+me+m—1=0. 

13. 2me’+a°+4a2+2mr+2m—4=0. 
14. 2mz’?+8mar—6=—824—2m— x’. 
15. m2’+9a2—-10=38ma—2277+2m. 


290 SCHOOL ALGEBRA. 


RELATIONS OF Roots AND COEFFICIENTS. 


302. Consider the equation z7—10x-+24=—0. Resolve 
into factors, (e —6)(2—4)=0. The two values of x are 
6 and 4; their sum is 10, the coefficient of x with its sign - 
changed; their product is 24, the third term. 


303, In general, representing the roots of the quadratic 
equation az*+ bz+-c=0 by 7, and 7, we have (§ 285), 


en be vO oe 

i, Sea eee ee 
20 2c 

Rate uth Oe en Aaa a 
¥ Qa 2 a 


Adding, Mtn =— : 3 


multiplying, 11, = 
If we divide the equation ax’ oh bz + ¢=0 through by 


a, we have the equation 2? + — “+2 =0; this may be 
C 


written x’?-+ px-+q¢=0 where 2, Gee 


It appears, then, that if any quadratic equation be made 
to assume the form 2?-+ px + q=0, the following relations 
hold between the coefficients and roots of the equation : 

(1) The sum of the two roots is equal to the coefficient 
of x with its sign changed. 

(2) The product of the two roots is equal to the constant 
term. 

Thus, the sum of the two roots of the equation 


v’—Tz+8=0 
is 7, and the product of the roots 8. 


PROPERTIES OF QUADRATICS. 291 


304. Resolution into Factors. By § 303, if 7, and 7 are 
the roots of the equation 2’-+ px-+q=0, the equation 
may be written 


Y?—(n+r)ctnr, = 0. 


The left member is the product of «—7, and x—7, so 
that the equation may be also written 


(7 — 17) (4 — 7.) = 0. 


It appears, then, that the factors of the guadratic expres- 
sion 2° + px-+q are x—71, and —*1,, where 7, and 7, are 
the roots of the quadratic equation «+ px+q=0. 

The factors are real and different, real and alike, or 
imaginary, according as 7, and 7, are real and unequal, 
real and equal, or imaginary. 

If 7, = 7, the equation becomes (x — 7) (a — 7) = 0, or 
(c—7,)?=0; if, then, the two roots of a quadratic equa- 
tion are equal, the left member, when all the terms are 
transposed to that member, will be a perfect square as 
regards x. 


305. If the equation is in the form az’+ ba -+e¢=0, the 
left member may be written 


(2° b C 
a\ x +2045), 
a(%—7,)(%#— 712). 


806. If the roots of a quadratic equation are given, we 
can readily form the equation. 


Ex. Form the equation of which the roots are 3 and — 2. 
The equation is (a — 3) (« + 7) == (), 


or (x — 3) (2% + 5) =0, 
or 2a? —x2x—-15=0, 


292 SCHOOL ALGEBRA. 


3807. Any quadratic expression may be resolved into 
factors by putting the expression equal to zero, and solving 
the equation thus formed. 


(1) Resolve into two factors 2? — 5x + 3. 


Write the equation 
a?—52e24+3=0. 


Solve this equation, and the roots are found to be 
r Fics 73 





. 


Therefore, the factors of «7 —5x2+3 are 
= bt V8 a SS VIS. 
2 2 
(2) Resolve into factors 327 — 42+ 5. 
Write the equation 
3a?—42+5=0. 
Solve this equation, and the roots are found to be 
2+ v—11 d 2—v—l11 
Therefore, the expression 327—4a+45 may be written (@ 305), 
3 ( 2+ vat) ( 2M It) 
ae POs Renee. ha. 


Exercise 104. 


Form the equations of which the roots are 


Lai beele, lt 9. 84 -V2,3— V2. 
2, 58, 6. — 1d = 120 910. ed eee 
Be Lee ee Lon oat ll. a,a—b. 

4. 4, 24. 8. =. 12. a+b,a—Bb. 
Resolve into factors, real or imaginary : 

13. 122°+ 24-1. 16. 2 —22+4+3. 

14. 827— 142 — 24, 17. 2tet+l. 


15. a? —27—2. 18. 2 —224 9, 


CHABTER XXL 
RATIO, PROPORTION, AND VARIATION. 


308. Ratio of Numbers. The relative magnitude of two 
numbers is called their ratio when expressed by the indi- 
cated quotient of the first by the second. Thus the ratio 


of a to b is or a+), or a:b. 


The first term of a ratio is called the antecedent, and the 
second term the consequent. When the antecedent is equal 
to the consequent, the ratio is called a ratio of equality ; 
when the antecedent is greater than the consequent, the 
ratio is called a ratio of greater inequality; when less, a ratio 
of less inequality. 

When the antecedent and consequent are interchanged, 
the resulting ratio is called the inverse of the given ratio. 
Thus, the ratio 3:6 is the znverse of the ratio 6: 8. 


809. A ratio will not be altered if both its terms are 
multiplied by the same number. For the ratio a:6 is 
Me , 
mb ’ 


represented by 7 the ratio ma: mb is represented by 


maa 
and since —-=-, we have ma:mb=a: 6. 
mb b 


310. A ratio will be altered if different multipliers of its 
terms are taken; and will be increased or diminished accord- 
ing as the multiplier of the antecedent is greater than or 
less than that of the consequent. Thus, 


994 SCHOOL ALGEBRA. 


If m>n, If m<n, 
then ma > na, then ma<na, 
d asd Si haat d dae pbs © 
sire nb ere se nb oy 
but LEIA but Mle 
: nb b i ibeEE 
Mma. a Mma a 
abe nb ~b 
or ma:nb>a:6b. or ma:nb<a:b. 


311. Ratios are compounded by taking the product of the 
fractions that represent them. Thus, the ratio compounded 
of a:b and ¢: dis ac: bd. 

The ratio compounded of a:6 and a:b is the duplicate 
ratio a: b?; the ratio compounded of a: 6, a:6, and a:b is 
the triplicate ratio a*: 6°; and so on. 


812. Ratios are compared by comparing the fractions 
that represent them. 


Thus, aonb > or ord, 
according as ae or < 7 
as “> or <7 
as ad> or < be. 


313. Proportion of Numbers. Four numbers, a, 6, c, d, are 
said to be in proportion when the ratio a: 6 is equal to the 
ratio ¢: d, 

We then write a:6=c:d, and read this, the ratio of a to 
& equals the ratio of ¢ to d, or a is to 6 as c is to d. 

A proportion is also written a:6::¢:d. 

The four numbers, a, 0, ¢, d, are called proportionals; a 
and d are called the extremes, 6 and c the means, 


RATIO AND PROPORTION. 295 


314, When four numbers are in proportion, the product 
of the extremes is equal to the product of the means. 


For, if Gets aot fe 
th ea 
en ae 
Multiplying by bd, ad=be. 
’ The equation ad=be gives a =“ wee so that an 
C 


extreme may be found by dividing the product of the 
means by the other extreme; and a mean may be found by 
dividing the product of the extremes by the other mean. 
If three terms of a proportion are given, it appears from 
the above that the fourth term can have one, and but one, 
value. 


315. If the product of two numbers is equal to the prod- 
uct of two others, either two may be made the extremes of 
a proportion and the other two the means. 


For, if ad = be, 


then, dividing by bd, nt is - 
or eee 
bd 
2 0 = 6" a: 


316. Transformations of a Proportion. If four numbers, a, 
6, ec, d, are in proportion, they will be in proportion by : 


I. Inversion; that is, b will be to a as dis toe. 


For, if C20; 
th a _£ 
en a 


296 SCHOOL ALGEBRA. 








nd shed A ee 
an ; - 
ee 
or -=-— 
Ge ce 
* 0 Oe. 
II. Composition; that is, a+ will be to 6 as e+distod. 
For, if Cie or 
then oS 
d oh ih) peat Le | 
an ; - at : 
Re atb_e¢e+-d 
b d 


0 ap aes XO: 


III. Division; that is, a—6 will be to 6 as e— ad i8 to d. 








For, if C200. a, 
th ie 
en ag 
and Rete te 
es C0 ee 
b d 


$0 — > Sb. 0 


IV. Composition and Division; that is, a+6 will be to 
a—obasc+distoc—d. 




















For, from II., Gah Oa od. 
b d 
and from III., Ose OE Cie, 
b d 
Dividing, Gotah peated, 
a—b c—d 
‘atb:a—b=cr+d:c-d. 


RATIO AND PROPORTION. 207 


V. Alternation; that is, a will be to ¢ as 6 is to d. 


Pore Geo. a, 
h po 
then Raa 

eee b ab be 

Multipl byes: gas 

ultiplying by : a, 
ad 
or SSS ee 
c hUd 

GC 0. d, 


317. In a series of equal ratios, the sum of the antecedents 
is to the sum of the consequents as any antecedent is to its 
consequent. 


Por sit 
y may be put for each of these ratios. 


Then rae =r, 2=7, Lay, 


Gj 
od 43 h 
(ad OF iC ree ite — he. 
s atetetg=(64+d4+f+A)r. 
RRA pg 
 b+d+fth b 
.atetetg:b+d+fth=a:b.- 


In like manner it may be shown that 


ma-+ne-+ pe+qg: mb+nd+pf+qh=a oD. 


318, QOontinued Proportion. Numbers are said to be in 
continued proportion when the first is to the second as the 
second is to the third, and so on. Thus, a, 8, c, d, are in 


continued proportion when 


g 
b 


298 SCHOOL ALGEBRA. 


819. Ifa, 5, c are proportionals, so that a:b =6:c, then 
6 is called a mean proportional between a and ec, and ¢ is 
called a third proportional to a and 6. 


If a:b6=b:c, then b= Vae. 
For, if Gio Oec, 
| me 
then ae 
and 100. 
b= Vac. 


320. The products of the corresponding terms of two or 
more proportions are in proportion. 


For, if Ot Dat, 
e:f=g:h, 
kal sr 1, 

th SSS I MS 

™ eV he 


Taking the product of the left members, and also of the 
right members of these equations, 


aek _cgm. 
bft dhn 
*, ack: bfl=cgm: dhn. 


321. Like powers, or like roots, of the terms of a propor- 
tion are in proportion. 


For, if 0b aa: 
then _ = : : 
Raising both sides to the nth power, 
ee 
piers he 


e a”: 6" =": d" 


RATIO AND PROPORTION. 299 


822. The laws that have been established for ratios 
should be remembered when ratios are expressed in frac- 
tional form. 

_ 2 

(1) Solve: ——— pots 

—a—-l] #+a- 

By composition and division, 

Lee Peg Br 
Q@+1) —2(%—2) 

This equation is satisfied when «=0. For any other value of 2, 

we may divide by 2. 


We then have 











and therefore 


(2) If a:6=c:d, show that 
J+tab:8—ab=2+ cd: d’?—cd. 











If ae 
then we 
and aia 
a Beam aay 
RAAT @+ab_c+ed cd 


Bab @—cd 
or a4+ab:b%?—ab=¢?+ cd: d*—cd. 


300 SCHOOL ALGEBRA. 


(3) If a:b =c:d, and a is the greatest term, show that 
a+d is greater than b+. 








Since “ = > and a> ¢, (1) 
the denominator de> it! 

From (1), by division, 2— ge oe a (2) 

b d 

Since b> Gd; 
from (2), a—b>c~—d. 

Now, — b4+d=b4d, 

Adding, at+d>b+e. 


323, Ratio of Quantities. To measwre a quantity of any 
kind is to find out how many times it contains another 
known quantity of the same kind, called the unit of measure. 
Thus, if a line contains 5 times the linear unit of measure, 


one yard, the length of the line is 5 yards. 


324, Commensurable Quantities. If two quantities of the 
same kind are so related that a unit of measure can be 
found which is contained in each of the quantities an in- 
tegral number of times, this unit of measure is a common 
measure of the two quantities, and the two quantities are 
said to be commensurable, 

If two commensurable quantities are measured by the 
same unit, their ratio is simply the ratio of the two numbers 
by which the quantities are expressed. Thus, + of a foot is 
a common measure of 24 feet and 32 feet, being contained 
in the first 15 times and in the second 22 times. 

The ratio of 24 feet to 32 feet is therefore the ratio of 
15 : 22. 

Evidently two quantities different in kind can have no 
ratio. 


RATIO AND PROPORTION. 301 


325. Incommensurable Quantities. The ratio of two quan- 
tities of the same kind cannot always be expressed by the 
ratio of two whole numbers. Thus, the side and diagonal 
of a square have no common measure; for, if the side is a 
inches long, the diagonal will be aV2 inches long, and no 
measure can be found which will be contained in each an 
integral number of times. 

Again, the diameter and circumference of a circle 
have no common measure, and are therefore incommen- 
surable. 

In this case, as there is no common measure of the two 
quantities, we cannot find their ratio by the method of 
§ 324. We therefore proceed as follows: 


Suppose a and 6 to be two incommensurable quantities 
of the same kind. Divide 6 into any integral number (7) 
of equal parts, and suppose one of these parts is contained in 
a more than m times and less thanmm-+1 times. Then the 





ratio {> but < Reels that is, the value of © lies 
eI: n b 


ome ag Teche 
n n | 
The error, therefore, in taking either of these values for 


between 





A is less than 1 But by increasing n indefinitely, 1 can 
n n 


be made to decrease indefinitely, and to become less than. 
any assigned value, however small, though it cannot be 
made absolutely equal to zero. 

Hence, the ratio of two incommensurable quantities can- 
not be expressed exactly in figures, but it may be expressed 
approximately to any desired degree of accuracy. 

Thus, if 6 represent the side of a square, and a the 
diagonal, 


= V2. 


MIR 


302 SCHOOL ALGEBRA. 


Now V2 = 1.41421356.....,a value greater than 1.414218, 
but less than 1.414214. 

If, then, a millionth part of b is taken as the unit, the 
1414213 Ke 1414214 
1000000 1000000’ 
and therefore differs from either of these fractions by less 








value of the ratio ;, lies between 


han ree isl 
1000000 
By carrying the decimal farther, a fraction may be found 
that will differ from the true value of the ratio by less than 
a billionth, a trilhonth, or any other assigned value what- 
ever. 


326. The ratio of two incommensurable quantities is an 
incommensurable ratio, and is a fixed value toward which 
its successive approximate values constantly tend as the 
error is made less and less. 


327. Proportion of Quantities. In order for four quanti- 
ties, A, B, C, D, to be in proportion, A and B must be of 
the same kind, and C and D of the same kind (but Cand 
D need not necessarily be of the same kind as A and B), 
and in addition the ratio of A to B must be equal to the 
ratio of C to D. 

If this be true, we have the proportion 


Aj B= ad): 


When four quantities are in proportion, the numbers by 
which they are expressed are four abstract numbers in 
proportion. 


828. The laws of $816, which apply to proportion of 
numbers, apply also to proportion of quantities, except that 
alternation will apply only when the four quantities in 
proportion are add of the same kind. 


RATIO AND PROPORTION. 303 


Exercise 105. 


1. Find the duplicate of the ratio 3: 4. 


2. Find the ratio compounded of the ratios 2: 3. 3:4, 
DP ne eee 
3. Find a third proportional to 21 and 28. 
Find a mean proportional between 6 and 24. 
Find a fourth proportional to 3, 5, and 42. 
Hine sat ow ll ee Bt 8. 
7. Find the number which must be added to both the 


terms of the ratio 3:5 in order that the resulting ratio 
may be equal to the ratio 15: 16. 


ao nr 


Ifa:b=c:d, show that 

Ssedce. bd = cya" 10. @—B:°-@P=a:e. 
9. ab : Cle 050" 11. 2a+0:2c+d=6:d. 
12. 5a—b:5e—d=a:e. 

13. a—3b:a+ 8b=c—3d:c+38d. 

14. a’ +ab+0':a0—ab+0?=—84+cd+a@:?—ced+d@. 
Find & in the proportion 

ie Gack de Bo erny eae Game S 

16 ae al 1Ste oe se Ad. 


19. Find two numbers in the ratio 2:3, the sum of whose 
squares is 325, 


20. Find two numbers in the ratio 5:38, the difference 
of whose squares is 400. 


21. Find three numbers Neh are to each other as 
2:3:5, such that half the sum of the greatest and least 
exceeds the other by 25. 


304 SCHOOL ALGEBRA. 


22. Find rif 6x—a:42—b=3824+6:2x4-+a. 
23. Find x and y from the proportionals 
2 y= eyes: ey a 0, 
24. Find w and y from the proportionals 
2eaty:y=sy:i2y—2; 
2¢+1:244+6=—y:y+2. 


a-tb-+e+d _a—b+e—ad how thine 
25: ie Pease Sg Sey eae ow tha 


VARIATION. 


329, A quantity which in any particular problem has a 
fixed value is called a constant quantity, or simply a constant; 
a quantity which may change its value is called a variable 
quantity, or simply a variable. 

Variable numbers, like unknown numbers, are generally 
represented by 2, y, 2, etc.; constant numbers, like known 
numbers, by a, 8, e, ete. 


330. Two variables may be so related that when a value 
of one is given, the corresponding value of the other can be 
found. In this case one variable is said to be a function | 
of the other; that is, one variable depends upon the other 
for its value. Thus, if the rate at which a man walks is 
known, the distance he walks can be found when the time 
is given ; the distance is in this case a function of the time. 


831, There is an unlimited number of ways in which 
two variables may be related. We shall consider in this 
chapter only a few of these ways. 


332, When 2 and y are so related that their ratio is 
constant, y is said to vary as x; this is abbreviated thus: 


“ ‘VARIATION. 805 


yoax. The sign o, called the sign of variation, is read 

“varies as.’’ Thus, the area of a triangle with a given 

base varies as its altitude; for, if the altitude is changed 

in any ratio, the area will be changed in the same ratio. 
In this case, if we represent the constant ratio by m, 


Rn We ese om; OAL 
Again, if y’, z’ and y", 2" be two sets of corresponding 
values of y and z, then 
Vee ST an 
or TER ee! Getia § 316, V. 


333, When z and yare so related that the ratio of y fae 
0 


is constant, y is said to vary inversely as x; this is written 
Yy © E Thus, the time required to do a certain amount of 
x 


work varies inversely as the number of workmen employed ; 
for, if the number of workmen be doubled, halved, or 
changed in any ratio, the time required will be halved, 
doubled, or changed in the inverse ratio. 


; i 
In this case, y Se EKO A Bee and zy=m; that is, 
the product xy is constant. 
1 1 
As before, y’ ae a a 
aly! = aly" 
or, Oe eae 7 § 315 


334, If the ratio of y: xz is constant, then y is said to 
vary jointly as x and z. 


In this case, 4 = mzz, 


and TT eaereetal = ahah: 


306 SCHOOL ALGEBRA. 


835, If the ratio y:~ is constant, then y varies directly 
z 


as 2 and inversely as z. 


max 
In this case, y=) 
Zz 
eel) " plas 
Tay Mele Ue, ade nce gee 
Ane ie ae ie ae het 
Zz Zz eo 12 


3836. Theorems. 


eb y cx, and x oc z, then y oz. 
For GT Pal ig, 
a) ie 
". y varies as Z. 
II. If you, and xz, then (y+z) oz. 
For y= mx and 2—= ne. 
.YtZ=(MEN)EZ; 
'. y+ z varies as &. 
Ill. If yox when z is constant, and yaz when 2 is 
constant, then y « xz when x and z are both variable. 
Let 2!, 7’, z', and 2", y", 2! be two sets of corresponding 
values of the variables. 
Let x change from 2! to x", zg remaining constant, and let 
the corresponding value of y be Y. 


Then af Ve oir ett (1) 
Now let z change from i to 2", 2 remaining constant. 
Then WBN ras Ait. (2) 
From (1) and (2), 
af Yon Y cole itt § 820 
or af ie Af et eas arcs | 
or panera arith ies Aha § 316, V. 


*. the ratio as constant, and y varies as xz. 


VARIATION. 307 


In like manner it may be shown that if y varies as each 
of any number of quantities x, z, u, etc., when the rest are 
unchanged, then when they all change, y « xzu, etc. Thus, 
the area of a rectangle varies as the base when the altitude 
is constant, and as the altitude when the base is constant, 
but as the product of the base and altitude when both vary. 


337, Examples. 


(1) If y varies inversely as x, and when y= 2 the cor- 
responding value of x is 386, find the corresponding value 
of x when y= 9. 


m 
Here =—, or m=ay. 
aw 


*-m=2X 36 = 72. 
If 9 and 72 be substituted for y and m respectively in 


x 
the result is 9= 2 or 9a = 72. 
x 
ge By Ans, 


(2) The weight of a sphere of given material varies as 
its volume, and its volume varies as the cube of its diam- 
eter. If a sphere 4 inches in diameter weighs 20 pounds, 
find the weight of a sphere 5 inches in diameter. 

Let W represent the weight, 


V represent the volume, 
D represent the diameter. 


Then WeV and Ve D3, 
. Wee Ds. 2 336, I. 
Put W=mD>; 
then, since 20 and 4 are corresponding values of W and D, 
20 = m x 64. 
20 5 
META 16 
* W=~, D. 


* when D=5. W=,5 of 125 = 39,4. 


308 SCHOOL ALGEBRA. 


Exercise 106. 
1. Ifwoy,and if y=8 when «=5, find x when yis 5. 


2. If W varies inversely as P, and W is 4 when P is 
15, find W when P is 12. 


3. If xoyand y cz, show that azay’. 
1 


4. lig coe and 4 bie show that x « z. 
Z 


5. If x varies inversely as y’—1, and is equal to 24 
when y= 10, find # when y =5. 


6. If x ‘varies as Peubs and is equal to 3 when y=1 
z 


and z= 2, show that zyz=2(y-+ 2). 


idee is 
7. If x—y varies inversely as Aan and #-+y varies 
inversely as Pee find the relation between x and z if 


2=1, y=8, when oe 


8. The area of a circle varies as the square-of its radius, 
and the area of a circle whose radius is 1 foot is 3.1416 
square feet. Find the area of a circle whose radius is 20 
feet. 


9. The volume of a sphere varies as the cube of its 
radius, and the volume of a sphere whose radius is 1 foot is 
4.188 cubic feet. Find the volume of a sphere whose radius 
is 2 feet. 


10. If a sphere of given material 3 inches in diameter 
weighs 24 lbs., how much will a sphere of the same material 
weigh if its diameter is 5 inches? 


VARIATION, 309 


11. The velocity of a falling body varies as the time 
during which it has fallen from rest. If the velocity of a 
falling body at the end of 2 seconds is 64 feet, what is its 
velocity at the end of 8 seconds? 


12. The distance a body falls from rest varies as the 
square of the time it is falling. If a body falls through 144 
feet in 8 seconds, how far will it fall in 5 seconds? 


The volume of a right circular cone varies jointly as 
its height and the square of the radius of its base. 

If the volume of a cone 7 feet high with a base whose 
radius is 8 feet is 66 cubic feet : 


13. Compare the volume of two cones, one of which is 
tivice as high as the other, but with one half its diameter. 


14. Find the volume of a cone 9 feet high with a base 
whose radius is 3 feet. 


15. Find the volume of a cone 7 feet high with a base 
whose radius is 4 feet. 


16. Find the volume of a cone 9 feet high with a base 
whose radius is 4 feet. 


17. The volume of a sphere varies as the cube of its 
radius. If the volume is 179% cubic feet when the radius 
is 34 feet, find the volume when the radius is 1 foot 6 
inches. 


18. Find the radius of a sphere whose volume is the sum 
of the volumes of two spheres with radu d+ feet and 6 feet 
respectively. 


19. The distance of the offing at sea varies as the square 
root of the height of the eye above the sea-level, and the 
distance is 8 miles when the height is 6 feet. Find the 
distance when the height is 24 feet. 


CHAPTER XXII. 


PROGRESSIONS. 


338, A succession of numbers that proceed according to 
some fixed law is called a series; the successive numbers are 
called the terms of the series. 

A series that ends at some particular term is a finite series; 
a series that continues without end is an infinite series. 


339, The number of different forms of series is unlimited ; 
in this chapter we shall consider only Arithmetical Series, 
Geometrical Series, and Harmonical Series. 


ARITHMETICAL PROGRESSION. 


340. A series is called an arithmetical series or an arith- 
metical progression when each succeeding term is obtained 
by adding to the preceding term a constant difference. 

The general representative of such a series will be 


a,a+d, a+t2d, at3d..... 
in which a is the first’ term and @ the common difference ; 


the series will be zncreasing or decreasing Pore as d 1s 
positive or negative. 


341. The nth Term. Since each succeeding term of the 
series 18 obtained by adding d to the preceding term, the 
coefficient of d will always be one less than the number of 
the term, so that the mth term is a+(n—1)d. 


If the nth term be represented by 7, we have 
l=a+(n—1)d. hs 


ARITHMETICAL PROGRESSION. OLE 


342, Sum of the Series. If 2 denotes the mth term, a the 
first term, 2 the number of terms, d the common difference, 


and s the sum of n terms, it is evident that 


! 


s= a +(atd)+(at+2d)+}--+0—d)+ 4, or 
s= 1 $(l—d)t(l—2d)$Hat a+ a, 
=n(a-+ f). 

s=2(a+l). II. 


343, mie the equations 
é=a+(n—1)d, i. 


s=T(a+), 1a. 


any two of the five numbers a, d, J, n, s, may be found when 
the other three are given. 


(1) Find the sum of ten terms of the series 2, 5, 8, 11... 


Here G= 2. G== 3, n =r10. 
From I., t=2+ 27 = 29, 
Substituting in IL., $= = (2 + 29) = 155. Ans. 


(2) The first term of an arithmetical series is 3, the last 
term 31, and the sum of the series 186. Find the series. 


From I., sl=34+(n—l)d. (1) 
From II., 136 = 530 +31). (2) 
From (2), pees 8. 


Substituting in (1), d= 4, 
he tseriesiis. 3. (rl barlo. LO oc OT) ol. 


512 SCHOOL ALGEBRA. 


(3) How many terms of the series 5, 9, 18, ....., must be 
taken in order that their sum may be 275? 
From I, l=5+(n—1)4. 
b=4n4T. (1) 
From II., O75 56 ry, (2) 


Substituting in (2) the value of / found in (1), 
275 = stn + 6), 
or 2n? + 3n = 275, 


This is a quadratic with n for the unknown number. 


Complete the square, 
16 n? +() +9 = 2209. 
Extract the root, 4n+3=+4 47. 


Therefore, »=11, or — 123. 


We use only the positive result. 


(4) Find x when d, J, s are given. 


From L., a=l—(n—1)d. 
From IL, qa 28am, 

n 
Therefore, l—~(n—1)d= 28 — In. 

n 


“. In—dn?+dn=2s—ln, 
“. dn? — (214+ d)n=— 2s. 


This is a quadratic with n for the unknown number. i 


Complete the square, 

4d°n? —() + (214+ dP =(21+ d)— 8ds. 
Extract the root, 
2dn—(214+d)=+4 V(214 d}— 8ds. 


21+d+V(21+ dy —8ds 
2d 


Therefore, n= 


ARITHMETICAL PROGRESSION. ole 
(5) Find the series in which the 15th term is — 25 and 
41st term — 41. 


If a is the first term and d the common difference, the 15th term 
is a + 14d, and the 41st term is a + 40d. 


Therefore, a+14d=— 25, (1) 
and a + 40d =— 41. (2) 
Subtracting, —26d= 16. ‘ 
Agee 
13 
Substituting in (1), a = — 16}. 


fae ee eee = Lo pipe 
Hence, the series is — 16,5,, —17, — 1745, ..... 


344. The arithmetical mean between two numbers is the 
number which stands between them, and makes with them 
an arithmetical series. 

If a and 6 represent two numbers, and A their arithmet- 
ical mean, then a, A, 6 are in arithmetical progression, and 
by the definition of an arithmetical series, § 340, 





A—a=d, 

and b6—A=d. 
. A—a=b—A 
Os ok 

) 


345. Sometimes it is required to insert several arithmeti- 
cal means between two numbers. 


Ex. Insert six arithmetical means between 8 and 17. 
Here the whole number of terms is eight; 3 is the first term and 
17 the eighth. 
By L,, pasa 7d. 
d= 2. 
_The series is 8, [5, 7, 9, 11, 18, 15,] 17, the terms in 
brackets being the means required, 


314 SCHOOL ALGEBRA. 


346, When the sum of a number of terms in arithmetical 
progression is given, it is convenient to represent the terms 
as follows: 


Three terms by x—y, 7, x+y; 
four terms by LOY, CY, oY, Oa Oy- 


and so on. 


Ex. The sum of three numbers in arithmetical progres- 
sion 1s 36, and the square of the mean exceeds the product 
of the two extremes by 49. Find the numbers. 


Let «—y, «x, «+ y represent the numbers. 
Then, adding, 3” = 36. 
* 2=12. 
Putting for « its value, the numbers are 
12—y, 12, 12+ y. 
Then (12)?—(12—y)(12 + y) = the excess. 


But 49 = the excess. 
Therefore, 144 — 144 + y? = 49. 
vy Yyesiee 7. 


Thesnumbersavaret5, 12) 19s. oreo sl2 0, 


Exercise 107. 


l=a+(n—l1)d; s==(a+1])=—[2a+ (n—1)d]. 


Find / and s, if 
1. Sed 4 et Se 


2. Os Did =5) es) 2a Detes 


ARITHMETICAL PROGRESSION. 315 


Find d and s, if 


(Ud wo 26 hosy een 
Sata On Ges 00 at Oe 
Ook i= LOU peer a7 ey 
TOD ea ee AD ape LE 


Insert eight arithmetical means between 


11. el etaniero: 13. 


12. aloes: 14. 


Find a ands, if 


TD ean) Pe 149 87 == 92). 16, 


Find 2 and s, if 


Un eb OU, 0 NO ma S. 


Find d and 2, if 


19° a= 11, (= 54, 6= 999.4, 20. 


Find d and J, if 


oo agi Aliso (ODUM 222 


Find a and d, if 


23. J=21,n=7,s=105. 24. 


Find a and J, if 


2b ceN== wl wa = 8 LIS eS Ge 


Find 7 and J, if 


OF ou ee OS0AG=49, == o) 28. 


Find a and 7, if 


29. s-=623,d=—5,l=77. 30. 


de 
2 4 


47 and 2. 

== 21 = 242 male, 
a= 34, (=10,d==2. 
a= 2, f= SO laga= SUL: 


==) 1) 2220 e305: 


t 


b= 10D, 2 == 1625 ==.840; 


1 
= 95, s=—75,d== 
nN s 9 


$==1(98)0 == 18, d= 6; 


e=1008 a= 44 = 88. 


316 SCHOOL ALGEBRA. 


Find the arithmetical series in which 


31. The 15th term is 25, and the 29th term 46. 


How many terms must be taken of 
32. The series —16, —15, —14,..... to make —100? 


33. The series 20, 183, 174, .....to make 1624? 


34. The sum of three numbers in arithmetical progres- 
sion is 9, and the sum of their squares is 29. Find the 
numbers. 


35. The sum of three numbers in arithmetical progres- 
sion is 12, and their product is 60. Find the numbers. 


GEOMETRICAL PROGRESSION. 


347, A series is called a geometrical series or a geometrical 
progression when each succeeding term is obtained by mul- 
tiplying the preceding term by a constant multiplier. 

The general representative of such a series will be 


OT Ot OT Ota 


in which a is the first term and ry the constant multiplier 
or ratio. 

The terms increase or decrease in numerical magnitude 
according as 7 is numerically greater than or numerically 
less than unity. 


348, The nth Term. Since the exponent of 7 increases by 
one for each succeeding term after the first, the exponent 
will always be one less than the number of the term, so 
that the nth term is ar"7}, 


If the nth term is represented by 7, we have 


Amey ete I, 


GEOMETRICAL PROGRESSION. Olt 


349, Sum of the Series. If / represent the mth term, a the 
first term, » the number of terms, r the common ratio, and 
s the sum of » terms, then 


s=a+t+ar Ze ar? eve 1) an (1) 
Multiply by 7, 
rs=artartartun ar"—' + ar, (2) 


Subtracting the first equation from the second, 


£§ — $= ar" — a, 





or oe saa 1), 
fee Cine) ie 
Pool 
Since = ar", r/= ar", and II. may be written 
rl— a 
= 0GE 
ee eg 


350, From the two equations I. and II., or the two equa- 
tions I. and III., any two of the five numbers a, 7, /, 7, s, 
may be found when the other three are given. 


(1) The first term of a geometrical series is 3, the last 
term 192, and the sum of the series 881. Find the number 
of terms and the ratio. 


From I., 192 37") (1) 
Broceti ie Sa Sent (2) 
r—1] 
From (2), p= 2. 
Substituting in (1), 2p Mes OF: 
o. sT. 


The series is 8, 6, 12, 24, 48 96, 192. 


318 SCHOOL ALGEBRA. 


(2) Find Z when », n, s are given. 








From L., a= be 
7r—-l 
ae s 
Substituting in ITI, $= ees 
r—l 
(r—1)s= Ge) 
pn—l 
rus (r — Sion 
yh — ] 


851, The geometrical mean between two numbers is the 
number which stands between them, and makes with them 
a geometrical series. 

If a and 6 denote two numbers, and G their geometrical 
mean, then a, G’, d are in geometrical progression, and by 
the definition of a geometrical series, § 347, 


= 7, and mat 
Mn Caan) 
pire heresies 
a Ga 


352. Sometimes it is required to insert several geometri- 
cal means between two numbers. 
Ex. Insert three geometrical means between 3 and 48. 


Here the whole number of terms is five; 3 is the first term and 48 
the fifth. 


By: 48 = 374, 
rt = 16. 
r=+ 2. 


The series is one of the following: 

2p Neh Pai ee Be 24), 48; 
3, [—6, 12, —24,] 48. 
The terms in brackets are the means required. 


’ 


GEOMETRICAL PROGRESSION. 319 


3538, Infinite Geometrical Series. When * is less than 1, the 
successive terms become numerically smaller and smaller ; 
by taking n large enough we can make the nth term, ay”“’, 
as small as we please, although we cannot make it absolutely 
zero. 














The sum of 7 terms, oa may be written j a -- a 
this sum differs from ae by the fraction oe by taking 


enough terms we can make /, and consequently the fraction 
rl 

Lz 4 

terms taken the nearer does their sum approach is 


,as small as we please; the greater the number of 








r 


Hence Pie is called the sum of an infinite number of 
9 


terms of the series. 


(1) Find the sum of the infinite series 1 — : -+- ' — | Se 


Here, a= 1, eens 
2 


The sum of the series is 





We find for the sum of n terms : — : = lies this sum evidently 


2 } 
approaches Se ee 


(2) Find the value of the recurring decimal 
0.12135185 ..... 


Consider first the part that recurs; this may be written 
135 
100000 


135 135 +..., and the sum of this series is ————_, 


100000 * 100000000 rT 
1000 


which reduces to a Adding 0.12, the part that does not recur, we 


obtain for the value of the decimal 12 ok ae O 449 


100" 740° ° 3700 





Ans. 


320 SCHOOL ALGEBRA. 


Exercise 108. 


b==.ay"—ls @ = = —— 


Find 7 and s, if 

Bie = Gee == 2) MA a=9,r=2, n= 11. 
Find r and s, if 

Sia 1) rash 6k 

Ati (it = Oy LOU: 

5. Insert 1 geometrical mean between 14 and 686. 

6. Insert 3 geometrical means between 31 and 496. 

Yu Mind -@ ands) 1b =a hoe 2 eee 

Sind ’s anda: alee == DOU ea 

9. Find r and n, if a=2, (= 1458, s= 2186. 


10. If the 5th term is = and the 7th term = find the 
series. 


11. Find three numbers in geometrical progression whose 
sum is 14, and the sum of whose squares 84. 


Sum to infinity : 


it at 2 2 jal 
190) eee ASO ee ee a OT bee ae 
2 ’ 9’ 4 ) "7 49 4 16 
13 eRa oka Epa oe 173 eee 
9 , 5 3 


Find the value of 
18. 0.16. 20. 0.86. 22. 0.736. 
19. 0.378. 21. 0.54. 23. 0.363. 


HARMONICAL PROGRESSION. B2l 


HARMONIGCAL PROGRESSION. 


354, A series is called a harmonical series, or a harmonical 
progression, when the reciprocals of its terms form an arith- 
metical series, - 

The general representative of such a series will be 

1 1 UAE GA Sa Sta a 
a atd at2d ‘ a+(n—l)d 

Questions relating to harmonical series are best solved 
by writing the reciprocals of its terms, and thus forming 
an arithmetical series. 








355. If a and 6 denote two numbers, and # their har- 
monical mean, then, by the definition of a harmonical series, 





nly Le el 

TT eas g 
Pe 
Wad 


» 856, Sometimes it is required to insert several harmoni- 
~ 
cal means between two numbers. 


Ex. Insert three harmonical means between 8 and 18. 








Find the three arithmetical means between 1 and i 
19 14 9 ean re. Se ‘ 
These are found to be 75! 79) 79) therefore, the harmonical means 
(pupa : 
e er vagh ho 343. 54, 8. 
O57, Since 4-2 -. Doe a eee VET cane wih 
a 


ae - dat Cee SALT: 


That is, the geometrical mean between two numbers is 
also the geometrical mean between the arithmetical and 
harmonical means of the numbers. 


B22 SCHOOL ALGEBRA. 


Exercise 109. 
1. If a, 6,¢ are in harmonical progression, show that 
a—b:b—c=a:e. 


2. Show that if the terms of a harmonical series are all 
multiplied by the same number, the products will form a 
harmonical progression. 


3. The second term of a harmonical series is 2, and the 
fourth term 6. Find the series. 


4. Insert the harmonical mean between 2 and 8. 


5. Insert 2 harmonical means between 1 and : 


6. Insert 5 harmonical means between 1 and 


7. The first term of a harmonical progression is 1, and 


the third term > Find the 8th term. 


8. The first term of a harmonical progression is 1, and 
the sum of the first three terms is 18. Find the series, 


9. Ifqa@is the arithmetical mean between 0 and ¢, and 8 
the geometrical mean between a and c, show that ¢ is the 
harmonical mean between a and 6. 


10. The arithmetical mean between two numbers exceeds 
the harmonical mean by 1, and twice the square of the 
arithmetical mean exceeds the sum of the squares of the 
harmonical and geometrical means by 11. Find the num- 
bers. 


CHAPTER XXIII. 
PROPERTIES OF SERIES. 


358, Convergent and Divergent Series. By performing the 
the 


infinite series 1+x2+2°+ 2°-+...... This series, however, 
is not equal to the fraction for all values of x. 





indicated division, we obtain from the fraction 


859, If is numerically less than 1, the series is equal to 
the fraction. In this case we can obtain an approximate 
value for the sum of the series by taking the sum of a 
number of terms; the greater the number of terms taken, 
the nearer will this approximate sum approach the value of © 
the fraction. The approximate sum will never be exactly 
equal to the fraction, however great the number of terms 
taken ; but by taking enough terms, it can be made to differ 
from the fraction as little as we please. 


Thus, if ate the value of the fraction is 2, and the 
series 18 


The sum of four terms of this series is 14; the sum of 
five terms, 113; the sum of six terms, 134; and so on. 
The successive approximate sums approach, but never 
reach, the finite value 2. 


360, An infinite series is said to be convergent when the 
sum of the terms, as the number of terms is indefinitely in- 
creased, approaches some fixed finite value; this finite value 
is called the sum of the series. 


324 SCHOOL ALGEBRA. 


3861. In the series 1 +ata?t at... suppose 2 
numerically greater than 1. In this case, the greater the 
number of terms taken, the greater will their sum be; by 
taking enough terms, we can make their sum as large as 
we please. The fraction, on the other hand, has a definite 
value. Hence, when x is numerically greater than 1, the 
series 1s not equal to the fraction. 

Thus, if «= 2, the value of the fraction is —1, and the 
series 18 


1t944tg4..... 


The greater the number of terms taken, the larger the sum. 
Evidently the fraction and the series are not equal. 


362. In the same series suppose 21. In this case the 


3 and the series 1+1+1-+1-..--. 


fraction is : 
1— 0 


The more terms we take, the greater will the sum of the 
series be, and the sum of the series does not approach a 
fixed finite value. 

If x, however, is not exactly 1, but is a little less than 1, 








the value of the fraction will be very great, and the 


x 
fraction will be equal to the series. 





Suppose «=—1. In this case the fraction is ; = ‘ — ; 
and the series 1—1+1—1+.=--.. If we take an even 


number of terms, their sum is 0; if an odd number, their 
sum is 1. Hence the fraction is not equal to the series. 


363. A series is said to be divergent when the sum of the 
terms, as the number of terms is indefinitely increased, 
either increases without end, or oscillates in value without 


approaching any fixed finite value. 


PROPERTIES OF SERIES. oo0 


No reasoning can be based on a divergent series; hence, 
in using an infinite series it 1s necessary to make such 
restrictions as will cause the series to be convergent. Thus, 
we can use the infinite series 1+a+a?+a2?+... when, 
and only when, z hes between +1 and —1. 


364, Identical Series. Jf two series, arranged by powers 
of x, are equal for all values of x that make both series con- 
vergent, the corresponding coefficients are equal each to each. 


For, if A+ Bo+ Ce? + eet = Alt Blat Ola? Beds 
by transposition, 


A— Al=(B'—B)xt+(C!—O)att 


? 


Now, by taking 2 sufficiently small, the right side of this 
equation can be made /ess than any assigned value what- 
ever, and therefore less than A—A', if A— A! have any 
value whatever. Hence, 4— A’ cannot have any value. 

Therefore, 

A— A'=0 or A= A’. 

Hence, Bz + Cz? + Da? +... = Bla + Ola? + D'aF +... 
or (B— B)2=(!—C)’+(D'—D) e+e; 
by dividing by z, 

B-B=(8—C)a#+(DI—-D)2 +; 
and, by the same proof as for A — A’, 
B- bi=- Vior.e = Bb" 
In like manner, 
(ee ae) ey 1 andeo O11. 
Hence, the equation 
At Bat Ce? foe Alt Bloat Chath wen, 
if true for all finite values of x, is an identical equation ; 
that is, (he coefficients of luke powers of x are the same. 


326 SCHOOL ALGEBRA. 


365, Indeterminate Coefficients. 


9 : 
Ex. Expand Dein SK ascending powers of 2. 


ltat2 
Assume Mem 5 + Bu + Cx? + Dz’ ..... , 
l+a+2? 


then, by clearing of fractions, 
24+32=A+ Bu + Cx? + Dad + + 
+ Av + Bu? + Cx8 + 0: 
+ Ax? + Ba + os 
“. 2430=A4(B4+A)x+(C+B+A)227+(D+ C+B) a3 +... 
By 3364, A=2, B+A=3, 0+ B+4+A=0, D+C+4+B=0; 
whence B=1, C=—3, D=2, and xo on. 


2+32 


elie Tega er ee py) 
wv Hb 


The series is of course equal to the fraction for only such values 
of « as make the series convergent. 


Nore. In employing the method of Indeterminate Coefficients, 
the form of the given expression must determine what powers of the 
variable z must be assumed. It 1s necessary and sufficient that the 
assumed equation, when simplified, shall have in the rght member 
all the powers of # that are found in the left member. 

If any powers of x occur in the right member that are not in the 
left member, the coefficients of these powers in the right member will 
vanish, so that 1n this case the method still applies; but if any powers 
of x occur in the left member that are not in the rght member, then 
the coefficients of these powers of « must be put equal to 0 in equating 
the coefficients of like powers of ; and this leads to absurd results. 
Thus, 1f 1t were assumed that 


24+32 


eyes ear 
x x 


there would be in the simplified equation no term on the right cor- 
responding to 2 on the left; so that, in equating the coefficients of 
like powers of a, 2, which 1s 22°, would have to be put equal to 02°; 
that 1s, 2=0, an absurdity. 


PROPERTIES OF SERIES. SH 


Exercise LIAO. 


Expand to four terms: 








peli as. Leas ie Yen mica 
1+ 2a ltat+2 l—xz—-27 
2. 1 7 5. BO Aw 8. REL siaty & 
2—32 lta2-2 ltate? 
3. Lia? 6. LoL 4 ae 1—82z 
2+32 1—22+ 82’ l—2—62 


Expand to five terms: 








10 4 d 12. Peo ee 14. ewie 2a) 
24g 1+382—27 a(x -~1) 
eats spre tn aia Sib Ee ant cals Wii 

3 + x(x — 2) (a—1)(2?+1) 


366. Partial Fractions. To resolve a fraction into partial 
fractions is to express it as the sum of a number of frac- 
tions of which the respective denominators are the factors 
of the denominator of the given fraction. This process is 
the reverse of the process of adding fractions which have 
different denominators. 

Resolution into partial fractions may be easily accom- 
plished by the use of indeterminate coefficients and the 
theorem of § 364. 

In decomposing a given fraction into its simplest partial 
fractions, it is important to determine what form the assumed 
fractions must have. 

Since the given fraction is the swm of the required par- 
tial fractions, each assumed denominator must be a factor 
of the given denominator; moreover, all the factors of the 
given denominator must be taken as denominators of the 
assumed fractions. 


328 SCHOOL ALGEBRA. 


Since the required partial fractions are to be in their 
simplest form incapable of further decomposition, the nu- 
merator of each required fraction must be assumed with 
reference to this condition. Thus, if the denominator 1s 
a” or (w+ a)”, the assumed fraction must be of the form 
A or a eA : for if it had the form A+ B or Art sb B 
a” . (+a) a (722 /a)* 
it could be decomposed into two fractions, and the partial 
fractions would not be in the simplest form possible. 

When all the monomial factors, and all the binomial 
factors, of the form w-:a, have been removed from the 
denominator of the given expression, there may remain 
quadratic factors which cannot be further resolved; and 
the numerators corresponding to these quadratic factors 
may each contain the first power of x, so that the assumed 


fractions must have either the form este or the 
Den avn s vrax+b 

form ==. 
vt b 





(1) Resolve 3 


j into partial fractions. 


Since a + 1=(e%+1)(a?—a +1), the denominators will be +1 
and #?—a+1. 





then 3 = A(a?—x2+1)+ (Be + C)(@ +1) 
=(A+ B)a?+(B+C—A)x+(A+C); 
whence, 3=A4C B+C—A=0, A+B=0, 
and A=1, B=—1, C=2, 
Therefore, : ale Bisa 








a ead e—aot+l 


PROPERTIES OF SERIES. 329 


Ji, oa Arey fe Mics , 
(2) Resolve Bes Ml = ODI into partial fractions. 
Bol 
The denominators may be 2, 27, +1, («+ 1). 
Meanie ee ee Oe a em ey Do 
x? (a + 1) Geese ware gr (etal)? 
*. 403-2? -32—2= Ax(e+1)?+ B(o+ 1)? 4+ Ce? (x +1) 4 Da? 
=(4+C)e+(2A4+B4+C+D)0e°+(A+2B)0+B; 














whence, A+C=4, 
2A4+B+C+D=-1, 
A+2B=—3, 
B=—2; 
or 1B ee ae Aiea Tee eg a ee ae 
rer aloe ee ee EON nO A ea: 
(a+ 1) aren re mien Loh (ates ¥)* 


Exercise 111. 


Resolve into partial fractions : 





(hee ae) Rist caiends ih 
(x +4) (a — 5) x’ (x +5) 
2 Tzx—l 8 if — 2 
(122) (1— 82) Gl @= 9) 
3 bol 5 yi ea tea aa, 
 (Qe—1)(@—5) bah tree Feo 
4 s— 2 10 (z—1 
 (@—5)(@ +2) » 6e+1)(2 +1) 
KE] ae 11. a er Outi 
xv—l (a+ 1) (@ + 2) 
v—x—3 12. v—a+l1 


OHAPTER XXIV. 
BINOMIAL THEOREM. 


867. Binomial Theorem, Positive Integral Exponent. By suc- 
cessive multiplication we obtain the following identities ; 

(a+b? =a@+2ab+ 0°; 

(a+ 6% = a + 80°) + 8a’ + 8; 

(a + 6) = at + 40°) + 60°? + 406? + 0. 

The expressions on the right may be written in a form 
better adapted to show the law of their formation : 


(a+ bY =a? + 2ab +228; 
ae | 
Wis 


Ai BAR aoe Mid Sora ay 
bse at 4a + att poe gph ae 
Ait 0) Bat 2) cat ae oe, 


Norr. The dot between the Arabic figures means the same as the 
sign X. 


be 





(a+ bf =a+3a7b 4S Sab} + 


368. Let represent the exponent of (a+ 6d) in any one 
of these identities; then, in the expressions on the right, 
we observe that the following laws hold true: 

I. The number of terms is n+ 1. 

II. The first term is a”, and the exponent of a is one 
less in each succeeding term. 

The first power of 4 occurs in the second term, the 
second power in the third term, and the exponent of 6 is 
one greater in each succeeding term. oF 

The sum of the exponents of a and 6 in any term is n. 


BINOMIAL THEOREM. 331 


III. The coefficient of the first term is 1; of the second 


term, 2; of the third term, aoe and so on, 


369. Consider the coefficient of any term; the number 
of factors in the numerator is the same as the number of 
factors in the denominator, and the number of factors in 
each is the same as the exponent of 4 in that term; this 
exponent is one less than the number of the term. 


370. Proof of the Theorem. That the laws of § 368 hold 
true when the exponent is any positive integer, is shown 
as follows : 

We know that the laws hold for the fourth power; 
suppose, for the moment, that they hold for the 4th power. 

We shall then have 


(a+ 6)* = ak + kak yp te k— 2h? 


Iago pa 
+ ah fee 1) 


Multiply both members of (1) by a+; the result is 
(a+ 6) al (h-+ Tath -{EEDE gry 
pGADEERD gripe pon 
In (1) put £+1 for &; this gives 
(a+ by = al + (k+4 lad + eerie YEFIAY pig 
Geb 41 -e+1— 2) age 4 


eoeee 


1°2°3 
ete, =a"t!+ (k+1) a*b + oS a? 6? 
Me ene 2B8 A ssse (3) 


Equation (3) 1s seen to be the same as equation (2). 


ban SCHOOL ALGEBRA. 


Hence (1) holds when we put &+1 for £; that is, if the 
laws of § 868 hold for the Ath power, they must hold for 
the (4 + 1)th power. 

But the laws hold for the fourth power; therefore they 
must hold for the fifth power. 

Holding for the fifth power, they must hold for the sixth 
power ; and'so on for any positive integral power. 

Therefore they must hold for the mth power, if 7 is a 
positive integer; and we have 
n a nw = 


(a + Wr — q” + na" + a” 26? 


n(n—1)(n— 


2) ag" —323 | wocce 
tie areas bet (A) 


Notr. The above proof is an example of a proof by mathematical 
unduction. 


371, This formula is known as the binomial theorem. 

The expression on the right is known as the expansion of 
(a+6)"; this expansion 1s a finite serves when 7 is a positive 
integer. That the series is finite may be seen as follows: 

In writing out the successive coefficients we shall finally 
arrive at a coefficient which contains the factor »—n; the 
corresponding term will vanish. The coefficients of all the 
succeeding terms likewise contain the factor n—n, and 
therefore all these terms will vanish. 


372, If a and 6 are interchanged, the identity (A) may 


be written 
Wee ee eg et n— n(n — 1) n—2 2 
(a+ 6)=(b+a)"=b"+ nb Ck, a 


n(n —1)(n = 2) n—58 ae 
T bee es: : i 


BINOMIAL THEOREM. Boo 


This last expansion is the expansion of (A) written in 
reverse order. Comparing the two expansions, we see 
that: the coefficient. of the last term is the same as the co- 
efficient of the first term; the coefficient of the last term 
but one is the same as the coefficient of the first term but 
one; and so on. 

In general, the coefficient of the 7th term from the end 
is the same as the coefficient of the rth term from the 
beginning. In writing out an expansion by the binomial 
theorem, after arriving at the middle term, we can shorten 
the work by observing that the remaining coefficients are 
those already found, taken in reverse order. 


373, If 6 is negative, the terms which involve even 
powers of 6 will be positive, and those which involve odd 
powers of 6 negative. Hence, 


(a — 6)" =a®— nab + pe ee “ cS 1) oe or 


ROSELL : 
[DBE RA ee A? 
Also, putting 1 for a@ and ~« for 0, in (A) and (B), 


(1ayt=14 e+ PG—dy 


4 min I Re ries (0) 


n(n —1) 
Lee 


2 


(l—2)*=1—ne+ 


SD Bite (D) 


334 SCHOOL ALGEBRA. 


374, Examples: 
(1) Expand (1+ 22) 
In (Q) put 2 for x and 5 forn. The result is 


5-449, 5:43 
1420) =145(20) +2 4402 4243 gas 
Slee cen) tne Vesey 

Hck VED cee 
[23:4 1-23-45 


==71 +102 + 40a? + 802? + 8024 + 3225, 


(2) Expand to three terms camer 
a 


Put a for 1 and 6 for a then, by (B), 
x 


(a — b) = a® — Babb + 15 atb? +»... 
Replacing a and 6 by their values, 


2-8) CF) BCE) 


2 
Te ge20 


375, Any Required Term. From (A) it is evident (§ 372) 
that the (7-+1)th term in the expansion of (a+ 0)” is 


EVAR Ae Mike ip 


Note. In finding the coefficient of the (r + 1)th term, write down 
the series of factors 1x 2x3.....r for the denominator of the co- 
efficient, then write over this series the factors n(n — 1)(n — 2), etc., 
writing just as many factors in the numerator as there are in the 
denominator. 


The (r+1)th term in the expansion of (a—)8)” is the 
same as the above if 7 is even, and the negative of the 
above if 7 is odd. 


BINOMIAL THEOREM. 335 


Ex, Find the eighth term of (4 = ae 


Here a=4, b=, Toe LOU nf, 
The term required is Lc Sg SNe 
b:+2°3°4°5f 6277 2 
which reduces to — 602i, 


376. A trinomial may be expanded by the binomial 
theorem as follows: 


Expand (1-+ 2% —2’)*. 
Put 22—a7=2: 
then (1 +23 =14 324 327 + 23, 
Replace z with 2a — a”. 
2. (1+ 2a — 28 =143(2a — 2%) + 3(2a — 2%)? + (2a — 2) 
=14+62+92? —423—9at + 625 — 26, 


Exercise 112. 








Expand : : 
1. (a+b). 6. (a? +8). 11. (=P 4+ ny 
: ate 7. (m+n) 19, (ant aby, 
3% bf oe 4 1 
( x ae 8. (a—d*)’. 13. (Qa? + 2)’, 
hs Ae 1 2 1 2 
a (5-4). 9 (a? o*) 14. (a? — 3)‘, 
B. (45-30) LOPS (Aine ee Tbh (2a? e/a) 
| a VON ava, Voi 
te) sa aia) 
17. (Fet4eva): Di GN Oe le PN 
gs Ve 2Na 


18. (2a°y! — yV yy. 21. (2ab? — bat)’. 


336 


22. 


24. 


SCHOOL ALGEBRA. 





aricy "a\° 2b bs 
a Me ea ry ial gee Wee 
ONS (; Ce 


Find the fourth term of (24 — 3y)'. 


) 


25. Find the ninety-seventh term of (2a —0)™. 


Nore. As the expansion has 101 terms, the ninety-seventh term 
from the beginning is the fifth term from the end. 


26. 


als 


28. 


29. 


30. 


31. 


32. 


33. 


34. 


35. 


36. 


37. 


38. 


39. 


40. 


41. 


Find the eighth term of (82 —y)". 
Find the tenth term of (2a? — 4a)”. 
Find the fifth term of (a— 2V2)*. 
Find the eleventh term of (2— a)". 
Find the fifteenth term of (x + y)”. 
Find the fourth term of (8 — 22)’. 


Find the twelfth term of (a?— aVz)". 


Find the seventh term of (y*— 1)®. 
Find the fifth term of (4a — bVb)". 


Find the fourth term of (Wa — V8)*. 
Find the third term of (Wa — V — 6)". 
Find the sixth term of (Wa? — V— 1)’. 


Find the eighth term of (V2a+ V3.2)”. 
Find the ninth term of («V-- 1+ yV— 1)®. 





; 3 fy -2\31 

Find the fifth term of («°6 a | . 
a 

Find the seventh term of (2+ 27)". 


BINOMIAL THEOREM. $37 


377. Binomial Theorem, Any Exponent. We have seen 
(§ 370) that when n is a positive integer we have the 
identity 


(1po)=1 na + MA) 1) ot NS 2) 3 ee 


We proceed to the case of fractional and negative expo- 
nents. 


I. Suppose x is a positive fraction, 2. We may assume 
that 
(1+ 2)? =(A+ Be + Ce? + Da? + «)f, (1) 
provided x be so taken that the series 
A+ Bu + Cv? + Da? + 
is convergent (§ 360). 
That this assumption is allowable may be seen as follows: 


Expand both members of (1). We obtain 


—] —] =%) 
pes en ODED. 





and A%’+qA*'Ba+ we (At Bit g APC) 274 ..... 


In the first & coefficients of the second series there enter 
only the first & of the coefficients A, B, C, D, ..... If, then, 
we equate the coefficients of corresponding terms in the 
two series (§ 364) as far as the /th term, we shall have just 
k equations to find k unknown numbers A, B, C, D, ..... 
Hence the assumption made in (1) is allowable. 

Comparing the two first terms and the two second terms, 
we obtain 


AN Soe BS Wed Fs 
Gat te por Gb = py it B=" 


338 SCHOOL ALGEBRA. 
Extracting the gth root of both members of (1), we have 


(Lt aja 14 Fat Oo + Dat + va (2) 


where x is to be so taken that the series on the right is 
convergent. 


II. Suppose 7 is a negative number, integral or frac- 
tional. Let »——™m, so that m is positive; then 
Ales Ps 1 
CE eee 
From (2), whether m is integral or fractional, we may 
assume 
AE Wanye aa te 
(lta2)" lt met ex?+ da + eee 


By actual division this gives an equation in the form 


(1+ 2)" =1— ma + Ce? + Da? + « (3) 


378. It appears from (2) and (8) that whether 7 be inte- 
gral or fractional, positive or negative, we may assume 


(1+ 2)*=1-+ ne t+ C2? + Da? +o, 
provided the series on the right is convergent. 
Squaring both members, - 
(l+22+2’?)"=1+2ne+2Ce?+ 2 Da? +... (1) 
+ nig? +2n02. 
Also, since 
I+ y*=1+ nyt CY + Dy +--+, 
we have, putting 22-+-2” for y, 
(1+22+2)"=1+n(2e2+27)4+C(22+2) 
ote D(2x -[- ae} een 
=1+2nr+ne* +4 C7 4+... 
+4 C2’+ 8 Dz’. (2) 


BINOMIAL THEOREM. 339 


Comparing corresponding coefficients in (1) and (2), 
n+40=20C+y7’, 
40+ 8D=2D+ 2nC 
*. 20=n?—n, and oan), 


8 D=(n—2)0, and D= eae 2 


and so on. 
Hence, whether be integral or fractional, positive or 
negative, we have 


(1+2)" "=1 net 2G) 1) 2 ESD) sebee ’ 





provided, always, x be so te that the series on the right 
is convergent. 

The series obtained will be an infinite series unless » is a 
positive integer (§ 371). 


379. If is negative, 


ee 1S — nar 2G su mu na Es 2) ieee 
Also, if <a, 


(a+2)" = (1 +3) 


if x >a, 
(a+2)"= @+ar=2(1 +) 


eee 


340 SCHOOL ALGEBRA. 


380, Examples. 
(1) Expand (1+ x), 


(+ajha14po+ Rb Dee 4 MARIE 


woeoe 





The above equation is only true for those values of « which make 
the series convergent. 


L 
(2) Expand Sara? 








meh 
1 ti 
=(1—2)7# 
V1i—2x 
HG 5 5e 
ay ni poe: stem WR eae See 
( Da ae a coe x” + 
1:5 12D¢0 
ales | 1 me Bee onc 
aT tee sie dais fae 


if « is so taken that the series is convergent. 
A root may often be extracted by means of an expansion. 


(3) Extract the cube root of 344 to six decimal places. 


1 1 
344 = 343/14 —\=78(14—}. 
( ee m( ae 


-. V344 = 7(1 Dare 


Waters 1h eh ae 
g Pe ed Pasa BAR Paibakn, 4 tt SEF! 
7( eal 1-2 Gale ) 


7(1 + 0,000971815 — 0.000000944), 


I 





= 7.006796. 
ot 
(4) Find the eighth term of Caae *. 
AN/y 
Here a=2, b= teenog Wee f== 7, 
4Vx 4a 2 
The term is ja sts ily Sirens teks vals Pa ay 
1:2°3°4:5°6°7 Ant 
3 1:3°B°7-9° 11-133 
2°4°6°8°10°12°14:47- gl 


BINOMIAL THEOREM. 341 


Exercise 113. 


Expand to four terms. 





~ (1+ x)? 6. (1+ x), 11. (Qa+ 3 ys, 
Mae ae areata 122-1 3y yas, 
mG 4a)? She (leas )s*: 13. 5 = 

: Vai—x 
. (l—2). Su baba) e. 
ete = yer S108) (Dp ba)%, V(a— 2x) 


15. 


16. 


nS 


ES. 


19. 


20. 


aL. 


22. 


23. 


24. 


25. 


26. 





Find the fourth term of (< ae ap 
QV x 


Find the fifth term of ——_. 
V(a — 2x) 


Find the third term of (4—72)?%.  * 
Find the sixth term of (a — anys. 


Find the fifth term of (1 —2a)7?. 


Find the fifth term of (1 — x)~*. 


Find the seventh term of (1 — x)?. 
1 
Find the third term of (1+ 2) 2, 


Find the fourth term of (1+ 2) ~ 8, 


Find the sixth term of ( i a 


x 


Find the fifth term of (24 —3y)~ 3 


Find the fourth term of (1 — 52)” 4 


CHAPTER XXV. 
LOGARITHMS. 


381. If the natural numbers are regarded as powers of 
ten, the exponents of the powers are the Common or Briggs 
Logarithms of the numbers. If A and B denote natural 
numbers, a and 6 their logarithms, then 10*= A, 10°= B; 
or, written in logarithmic form, log A =a, log B= b. 


382. The logarithm of a product is found by adding the 
logarithms of its factors. 


For AC Be GU sa Ny Oe 


cS ] 


Therefore, log(A x B)=a+60=log A+ log B. 


383, The logarithm of a quotient is found by subtracting 
the logarithm of the divisor from that of the dividend. 


aS 10e 
F ——-—= 10%". 
B10 
ae 
Therefore, log ep b= log A — log B 


384, The logarithm of a power of a number is found by 
multiplying the logarithm of the number by the exponent 
of the power. 


For Aries CLS ies LNs 
Therefore, log A"*=an=n log A. 


LOGARITHMS. 343 


885. The logarithm of the root of a number is Pa by 
dividing the logarithm of the number by the index of the 
root. 


For YA = VIO = 10" 
Therefore, log VA = a, — 


386, The logarithms of 1, 10, 100, etc., and of 0.1, 0.01, 
0.001, etc., are integral numbers. The logarithms of all 
other numbers are fractions. 


Since 10? — "1; LOG? (at each, 
et Ot 10? (=74,) =0.01, 
10? = 100, 10-° (= yp) = 0.001, 
therefore log 1=—0, log0.1 =—I, 
log 10= 1, log 0.01 =— 2, 
log 100 = 2, log 0.001 = — 3. 


Also, it is evident that the common logarithms of all 
numbers between 

land 10will be 0O-+ a fraction, 

10 and 100 willbe 1- a fraction, 

100 and 1000 will be 2+ a fraction, 

land0.1 will be —1-+a fraction, 

0.1 and 0.01 will be —2-+a fraction, 

0.01 and 0.001 will be — 3+ a fraction. 


387. If the number is less than 1, the logarithm is nega- 
tive (§ 386), but is written in such a form that the fractional 
part is always positive. 


388, Every logarithm, therefore, consists of two parts: a 
positive or negative integral number, which is called the 
characteristic, and a positive proper fraction, which is called 


344 SCHOOL ALGEBRA. 

e 
the mantissa. Thus, in the logarithm 3.5218, the integral 
number 3 is the characteristic, and the fraction .5218 the 
mantissa. In the logarithm 0.7825 — 2, which is sometimes 
written 2.7825, the integral number — 2 is the character- 
istic, and the fraction .7825 is the mantissa. 


389, If the logarithm has a negative characteristic, it is 
customary to change its form by adding 10, or a multiple 
of 10, to the characteristic, and then indicating the sub- 
traction of the same number from the result. Thus, the 
logarithm 2.7825 is changed to 8.7825—10 by adding 10 to 
the characteristic and writing —10 after the result. The 
logarithm 13.9278 is changed to 7.9273—20 by adding 20 
to the characteristic and writing — 20 after the result. 


390. The following rules are derived from § 386: 


RuuE 1. If the number is greater than 1, make the 
characteristic of the logarithm one unit less than the num- 
ber of figures on the left of the decimal point. 

RuueE 2. If the number is less than 1, make the charac- 
teristic of the logarithm negétive, and one unt more than 
the number of zeros between the decimal point and the first 
significant figure of the given number. 

Rue 8. If the characteristic of a given logarithm is 
positive, make the number of figures in the integral part of 
the corresponding number one more than the number of 
units in the characteristic. 

Ruuz 4. If the characteristic is negatwe, make the num- 
ber of zeros between the decimal point and the first sig- 
nificant figure of the corresponding number one Jdess than 
the number of units in the characteristic. 


Thus, the characteristic of log 7849.27 is 38; the character- 
istic of log 0.037 is —2=8.0000—10. If the characteristic 


LOGARITHMS. 345 


is 4, the corresponding number has five figures in its integral 
part. If the characteristic 1s — 3, that is, 7.0000 — 10, the 
corresponding fraction has two zeros between the decimal 
point and the first significant figure. 


391, The mantissa of the common logarithm of any inte- 
gral number, or decimal fraction, depends only upon the 
digits of the number, and is unchanged so long as the 
sequence of the digits remains the same. 

For changing the position of the decimal point in a 
number is equivalent to multiplying or dividing the num- 
ber by a power of 10. Its common logarithm, therefore, 
will be increased or diminished by the exponent of that 
power of 10; and since this exponent is integral, the man- 
tussa, or decimal part of the logarithm, will be unaffected. 


Thus, 27196 = 10+, 2.7196 = 10%, 
NALIN fem g id Td 0.27196 = 10%#45-10 
27.196 = 10™,  0.0027196 = 1074-10, 


One advantage of using the number ten as the base of a 
system of logarithms consists in the fact that the mantissa 
depends only on the sequence of digits, and the characteristic 
on the position of the decimal pornt. 


392, In simplifying the logarithm of a root the equal 
positive and negative numbers to be added to the logarithm 
should be such that the resulting negative number, when 
divided by the index of the root, gives a quotient of — 10. 

Thus, if the log 0.0023 = 4 of (7.8010 — 10), the expres- 
sion 4 of (7.3010 — 10) may be put in the form 4 of 
(27.3010 — 30), which is 9.1003 — 10, since the addition of 
20 to the 7, and of — 20 to the — 10, produces no change 
in the value of the logarithm. 


346 SCHOOL ALGEBRA. 


Exercise 114. 


Given: log 2=0.38010; log3=0.4771; log5=0.6990 ; 
log 7= 0.8451. 

Find the common logarithms of the following numbers 
by resolving the numbers into factors, and taking the sum 
of the logarithms of the factors: 


It log bo} be loeiaoe 9) “log 0.0210 1S loots 
2. log 15. 6. log 80. 10. log 0.35. 14. log 16. 

3. log 21,7. log 42.4, 116 log 0.0035... 15. log.0.056: 
4. log 14. 8. log 420. 12. log 0.004. 16. log 0.63. 


Find the common logarithms of the following: 


vi 
1758 20) (bi Gases wee 26. 77. 29. 5%. 
18)..f'~, Q1eO!. ae Ode eo Te oe 30. 27. 
19:8. 74) Ub ooUess 4 oon mma pe)! Qirehey be 


393. The logarithm of the reciprocal of a number is 
called the cologarithm of the number. 
If A denote any number, then 


colog. A = log = = log1 — log A (§ 3883) = — log A (§ 386). 


Hence, the cologarithm of a number is equal to the log- 
arithm of the number with the minus sign prefixed, which 
sign affects the entire logarithm, both characteristic and 
mantissa. 

In order to avoid a negative mantissa in the cologarithm, 
it is customary to substitute for — log A its equivalent 
(10 — log A) — 10. 

Hence, the cologarithm of a number is found by subtract- 
ing the logarithm of the number from 10, and then annexing 
— 10 to the remainder. 


LOGARITHMS. 347 


The best way to perform the subtraction is to begin on 
the left and subtract each figure of log A from 9 until we 
reach the last significant figure, which must be subtracted 
from 10. 

If log A is greater in absolute value than 10 and less 
than 20, then in order to avoid a negative mantissa, it is 
necessary to write —log A in the form (20 — log A) — 20. 
So that, in this case, colog A is found by subtracting log A 
from 20, and then annexing — 20 to the remainder. 


(1) Find the cologarithm of 4007. 


10 —10 
Given: log 4007 = 3.6028 
Therefore, colog 4007 = 6.3972 — 10 


(2) Find the cologarithm of 103992000000. 


20 — 20 
Given : log 103992000000 = 11.0170 


Therefore, colog 103992000000 = 8.9830 — 20 


If the characteristic of log A is negative, then the subtra- 
hend, —10 or —20, will vanish in finding the value of 
colog A. 


(3) Find the cologarithm of 0.004007. 


10 —10 
Given: log 0.004007 = 7.6028 — 10 
Therefore, colog 0.004007 = 2.3972 


By using cologarithms the inconvenience of subtracting the 
logarithm of a divisor is avoided. For dividing by a num- 
ber is equivalent to multiplying by its reciprocal» Hence, 
instead of subtracting the logarithm of a divisor, its colog- 


arithm may be added. 


348 SCHOOL ALGEBRA. 





4) Find the logarithm of 
(4) Find the logarithm o 6.003 


log ae = log 5 + colog 0.002. 
log 5 = 0.6990 
colog 0.002 = 2.6990 


log quotient = 3.3980 
0.07 
pie 


log “2 = log 0.07 + colog 2°. 





(5) Find the logarithm of 


log 0.07 = 8.8451 — 10 
colog 2? = (10 — 3 log 2) — 10 = 9.0970 — 10 
log quotient = 7.9421 — 10 


Exercise 115. 


Given: log 2=0.38010; log 3=0.4771; log 5= 0.6990; 
log 7 = 0.8451; log 11 = 1.0414. 
Find the logarithms of the following quotients: 





he saps 15, 0.08. a2, &. 
if 1g 44 a: 
2. £. Hills 16. 4. 93, 2%. 
5 iti 0.05 28? 
pce 10. 2. 17. 22. gals 
6 8 Geld 15% 
ae 1 is, 24. ee eas 
D th Do 0.005 
5. ie 12. a 19. it’ 26. 0.007, 
ib 22 Me 15° 
imE 0% 3 0.011 
Ppa cer pe eake, 202. eh sit ato 
7 ge 6* As 0.05% 
5 6.05 he 0.05% 
17 ies 14e Pie 28. ! 
7 4 2° 0.011 


LOGARITHMS. 349 


394, Tables. A table of four-place common logarithms 
is given at the end of this chapter, which contains the com- 
mon logarithms of all numbers under 1000, the decimal pownt 
and characteristic being omitted. The logarithms of single 
digits, 1, 8, etc., will be found at 10, 80, etc. 

Tables containing logarithms of more places can be pro- 
cured, but this table will serve for many practical uses, and 
will enable the student to use tables of five-place, seven- 
place, and ten-place logarithms, in work that requires 
greater accuracy. 

In working with a four-place table, the numbers corre- 
sponding to the logarithms, that is, the antilogarithms, as 
they are called, may be carried to fowr segnificant digits. 


395. To Find the Logarithm of a Number in this Table. 


(1) Suppose it is required to find the logarithm of 65.7. 
In the column headed ‘‘N”’ look for the first two significant 
figures, and at the top of the table for the third significant 
figure. In the line with 65, and in the column headed 7, 
is seen 8176. To this number prefix the characteristic and 
insert the decimal point. Thus, 


log 65.7 = 1.8176. 


(2) Suppose it is required to find the logarithm of 20347. 
In the line with 20, and in the column headed 8, is seen 
3075; also in the line with 20, and in the 4 column, is seen 
3096, and the difference between these two is 21. The dif- 
ference between 20300 and 20400 is 100, and the difference 
between 20300 and 20347 is 47. Hence, 74% of 21 = 10, 
nearly, must be added to 8075; that is, 


log 20347 = 4.3085. 
(3) Suppose it is required to find the logarithm of 


0.0005076. In the line with 50, and in the 7 column, is 
seen 7050; in the 8 column, 7059: the difference is 9. The 


350 SCHOOL ALGEBRA. 


difference between 5070 and 5080 is 10, and the difference 
between 5070 and 5076 is 6. Hence, 3% of 9=5 must be 
added to 7050; that is, 


log 0.0005076 = 6.7055 — 10. 


396. To Find a Number when its Logarithm is Given. 

(1) Suppose it is required to find the number of which 
the logarithm is 1.9736. 

Look for 9736 in the table. In the column headed “N,”’ 
and in the line with 9736, is seen 94, and at the head of 
the column in which 9786 stands is seen 1. Therefore, 
write 941, and insert the decimal point as the characteristic 
directs; that is, the number required is 94.1. 


(2) Suppose it is required to find the number of which 
the logarithm is 3.7936. 

Look for 7936 in the table. It cannot be found, but the 
two adjacent mantissas between which it lies are seen to be 
7931 and 7988; their difference is 7, and the difference be- 
tween 7931 and 7936 is 5. Therefore, $ of the difference 
between the numbers corresponding to the mantissas, 7931 
and 7938, must be added to the number corresponding to 
the mantissa 7931. 

The number corresponding to the mantissa 7938 is 6220. 

The number corresponding to the mantissa 7931 is 6210. 

The difference between these numbers is 10, 
and 6210 + $ of 10 = 6217. 

Therefore, the number required is 6217. 


(3) Suppose it is required to find the number of which 
the logarithm is 7.38882 — 10. 

Look for 8882 in the table. It cannot be found, but the 
two adjacent mantissas between which it lies are seen to be 
38874 and 8892; the difference between the two mantissas 
is 18, and the difference between 3874 and the given man- 
tissa 3882 is 8. 


LOGARITHMS. 85h 


The number corresponding to the mantissa 3892 is 2450. 
The number corresponding to the mantissa 3874 is 2440. 
The difference between these numbers is 10, 

and 2440 + 58 of 10 = 2444. 
Therefore, the number required is 0.002444. 


Exercise 116. 


Find, from the table, the common logarithms of: 

IU! 4. 7808. Te 1063, 10. 000.5234. 
de AON 5. 4825. seg ae AD De 11. 0,01423: 
3. 888. 6. 8109. 9.70 O07 80m 5 9.132). 01987; 


Find antilogarithms to the following common logarithms: 
13. 4, 1432) Th ae ok 17. 9.0380 — 10. 
14 fo 3.00 kee 16.) 1.3709: 18. 9.9204 — 10. 


397, Examples. 


(1) Find the product of 908.4 x 0.05392 x 2.117. 


log 908.4 = 2.9583 
log 0.05392 = 8.7318 — 10 
log” 2417 = 0.3257 
2.0158 = log 103.7. Ans. 

When any of the factors are negative, find their logarithms with- 
out regard to the signs; write — after the logarithm that corresponds 
to a negative number. If the number of logarithms so marked is 
odd, the product is negative; if even, the product is positive. 


— 8.3709 x 884.637 
7308.946 
log 8.3709=0,9227  — 
log 834.637=2.9215 + 
colog 7308.946 = 6.1362 — 10 + 


9.9804 — 10 = log — 0.9558. Ans. 


(2) Find the quotient of 


352 SCHOOL ALGEBRA. 


(3) Find the cube of 0.0497. 
log 0.0497 = 8.6964 — 10 
Multiply by 3, 3 
6.0892 — 10 = log 0.0001228. Ans. 


(4) Find the fourth root of 0.00862. 
log 0.00862 = 7.9355 — 10 
Add 30 — 30, 30. — 30 
Divide by 4, 4) 37.9355 — 40 
9.4839 — 10 = log 0.3047. Ans. 


30.103* x 0.43482 x 69.897! 


log 3.1416 = 0.4971 = 0.4971 
log 4771.21—= 3.6786 = 3.6786 
Llog 2.7188 = 0.4848+2 =0.2172 
4colog 30.103 = 4(8.5214 — 10) = 4.0856 — 10 
Lcolog 0.4343 = 0.3622+2 =0,1811 
4colog 69.897 = 4(8.1555 — 10) = 2.6220 — 10 
11.2816 — 20 
S080 


8.2563 — 10 
= log 0.01804. Ans. 


(5) Find the value of ‘| 


398. An exponential equation, that is, an equation in which 
the exponent involves the unknown number, is easily solved 


by Logarithms. 


Ex. Find the value of x in 81*= 10. 
812 = 10. 
 . log (81*) = log 10, 
x log 81 = log 10, 


gy = 108.10 _ 1.0000 _ 9 504 Ans. 
log 81 1.9085 


LOGARITHMS. 353 


Exercise 117. 


Find by logarithms the following products : 


1. 948.7 x 0.04387. 5. 7564 x (— 0.003764). 
2. 3.409 X 0.008763. 6. 3.764 x (— 0.08349). 
3. 830.7 x 0.0003769. 72 — 0.040 « (= (00178). 
4. 8.439 x 0.9827. 8. — 8945.7 x 73.84. 


Find by logarithms: 


7065. a 0.07654 
* 5401 ' 83.94 x 0.8395 
a es 1p, 212% (—6.12) x (—2008) 
* — 0.06875 N9365 X (581) x 2.576 
13. 0.1768". 17. 44), 21. 2.56321, 
14. 1.211° 18. 906.84, 22. (88)%, 
15. 114. 19. (284)°, 23. (581), 
16. (78)", 20. (Ty8)™. 24. (920)>, 


25. 0.0075" x 78.34 x 172.4% x 0.00052 
42853 x 54.274 x 0.001 x 86.792 


26, (0.08271? x 8.429 x 0.7752" 
32.79 x 0.000371' 


27. | 7.126 x 0.1327 x 0.05738 _ 
0.4346 x 17.38 x 0.006372 


Find x from the equations : 
eon) 1.0. S07 2-0), 82. (0.4) = 3: 
OO ar c= Ah) 31. (1.8)? =42, 33. (0.9)*=2, 


304 SCHOOL ALGEBRA. 
NivO [a veuiee 


10 43 | 0086 | 0128 
0492 | 0531 
0864 | 0899 
1206 ; 1239 
1523 | 1553 


1818 | 1847 
2095 | 2122 
2355 | 2380 
2601 | 2625 
2833 | 2856 


3054 | 3075 
3263 | 3284 
3464 | 3483 
3655 | 3674 
3838 3856 


4|5 


0170 | 0212 
0569 | 0607 
0934 | 0969 
1271 | 1303 
1584 | 1614 


1875 |. 1903 
2148 | 2175 
2405 | 2430 
2648 | 2672 
2878 | 2900 


3096 | 3118 
3304 | 3324 
3502 | 3522 
3692 | 3711 
3874 | 3892 





4216 | 4232 
4378 | 4393 
4533 | 4548 
4683 | 4698 


4829 | 4843 
4969 | 4983 
5105 | 5119 
5237 | 5250 
5366 | 5378 


5490 | 5502 
5611 | 5623 
5729 | 5740 
5843 | 5855 
5955 | 5966 


6064 | 6075 
6170 | 6180 
6274 | 6284 
6375 | 6385 
6474 | 6484 


6571 | 6580 
6665 | 6675 
6758 | 6767 
6848 | 6857 
6937 | 6946 6964 | 6972 | 6 


7024 | 7033 7050 | 7059 
7110} 7118 7135 | 7143 
7193 | 7202 7218 | 7226 
7275 | 7284 7300 | 7308 
7356 | 7364 2 | 7380 | 7388 


4183 | 4200 
4346 | 4362 
4502 | 4518 
4654 | 4669 


4800 | 4814 
4942 | 4955 
5079 | 5092 
5211 | 5224 
5340 | 5353 


5465 | 5478 
5587 | 5599 
5705 | 5717 
5821 | 5832 
5933 | 5944 


6042 | 6053 
6149 | 6160 
6253 | 6263 
6355 | 6365 
6454 | 6464 


6551 | 6561 
6646 | 6656 
6739 | 6749 
6830 | 6839 
6920 | 6928 


7007 | 7016 
7093 | 7101 
7177 | 7185 
7259 | 7267 
7340 | 7348 








Fee 
eee 
LEEESe 
eooge 
pease 
Beeee 
caegs 
aac 











LOGARITHMS. 


7466 
7543 
7619 
7694 
7767 


7839 
7910 
7980 
8048 
8116 


8182 
8248 
8312 
8376 
8439 


8500 
8561 
8621 
8681 
8739 


8797 
8854 
8910 
8965 
9020 


9074 
9128 
9180 
9232 
9284 








9335 
9385 
9435 
9484 
9533 | 9538 





ee 
geeee 
Beeee 
Benes 
ones 
BEees 
= 
aeaee 
EGES 


9581 | 9586 
9628 
9675 
9722 
9768 


9814 
9859 
9903 
9948 
9991 


855 





CHAPTER XXVI. 


GENERAL REVIEW EXERCISE. 


If a=6, b=5, e=—4, d=-— 83, find the value of . 


ik 


5. 


V6 + ac Ve — ae. 3. Ve? + ac-+ Ve —2ae. 





Oi Btn ako eS 

2a—V8?—ae ce + 2d(d? — c’) 
Find the value of 

x abe 

a Fed tan h = : 

The when 2 eee 

1 pies) Ge nt Sten ca eee 

a b a a 

x a’(b — a) 

aug S Boal ad h = 

PAR RE Mee ge tama 


10. 


Hy 


12. 


13. 


C 


(a+2)(6+2)—a(b6+c)+2’, when Baars 


aQl+o)+be a , when 2=—a. 
a(l+6)—6be¢ a—2bez 


Add (a—6)2#’+(6—c)y’+(c—a) 2’, (6—¢) 2+ (e—a)y’ 
+ (a — b)2’, and (e—a)2’?+ (a— b)y?+(b6—e)2’. 
Add (a+6)a2+(b+e)y—(e+a)z, (64+e)z2+(e+a)a 
—(a+b)y, and (ate)y+(a+6)z2—(b+e)z2. 
Show that 2° -+-7°+2—32yz=0,if2+ty+z=0. — 
Show that 2°— 87°— 272— 18xyz=0, if x= 2y+3z. 





GENERAL REVIEW EXERCISE. 357 


Simplify by removing parenthesis and collecting terms: 
14. 83a—2(b—c)—[2(a—b)—3(e+a)]—[9e—4(e—a)]. 
5. 7(2a+ 6) —§19b—[18(e—a)+12(6—c)]}. 

16. «—j4y+[8(¢—2)—(#+ 2y)|— Qy+2—22)}. 
17. 14+ 2§2+4+4—3[24+5—4(¢+1)}}. 

18. 10% —§4[5a¢— 3(¢—1)]—3[42—3(¢+1)}}. 

19. 3a°—§22?—(82—7)—[2 2’— (38 2—2’) |—[5—(2 2°42) }}. 
20. (v7 — 2) (@ — 3) — (a — 7) (#— 1) + (e#— 1) (@ — 2). 
21. (2+ y)— 2" (32+ 2y)—(y—2)(—a2+y). 

22. 3a— [2a — (Ba BY] + 8a(20 30-5) 


a 


Resolve into lowest factors: 


23. (a+yy—42. 27. 922 — Baty? + by, 
24. (x? +4)? — 42°47. 28. (ies (lee oy 
25. a — B?— + Qe. 29. 8la*—l. 
26. (2? yr —2y— Ay’. 30. a®—b", 
31. (a — 0+ e —d’)Y — (2ac — 2bd). 
32. 2 —19x+ 84. 40. v—y¥*. 
33. 427+ 2ha— 36. 41. 8+ ax’. 
34. 2?— 824+ 15. 42. 2° — q”. 
35. 927 — 1502+ 600. 43. 27a°® — 64. 
36. 562°+ 82y — 207’. 44, x" — 827. 
37. 122° + 372-4 21. 45. a”®— 6. 
38. 83 — 142 — 402”. 46. 2° + 1024y"” 
89. 6a°+ 52 —4. 47. a —(b+c)’. 


48. 82°— 6ay(227+ 3y) 4 27y’. 


358 SCHOOL ALGEBRA. 


Find the H.C.F. of 
49. 62*— 22° 4+ 927+92—4, and 92*+ 8027 — 9. 
50. 82° — 52° +2, and 22°—52’? +3. 
51. 2° — 932 — 308, and 2° — 2127? + 1814 — 231, 
52. 2 —2e°+ 42?— 62+ 8, and a*— 22°— 227+ 64 —38. 
538. 2 —42°— 27 +2¢742 and 2? — 2? —2a24 2. 
54. 382°+102°+ Tx — 2, and 82°+13827+17¢+6. 
55. 4a*—92°+6x2—1, and 62°—72’?+1. 
56. «° +1la2—12, and 2°+ lla’+ 54. 


Find the L.C.M. of 
57. 42°+4a2—8, and 427+ 2x2—6. 
58. xv’? —4y’, and 2’? + 2y— by’. 
59. Ta’x(a—x), 2lar(a?— x’), and 12az*(a+2). 
60. 9a°—a2—2, and 32°— 102? — Tx — 4. 
61. 2?—527+6, 2?—4274+3, and 2?— 824 2. 
62. 2a + Say — bay*?+y*, and 22°— Txy+ day’?- y’. 
Simplify : | 
Pere in ae Sees Sie aoe 
TAC == Dye (ey) 
lt+e¢,l—2z l1+2? -l-“# : 





63. 











64. — 

l—« lte 1-2 1l+2# 
ek Cel _ 2(e@ +2) Lis z+5 
 (@+2)(¢@+5) (@+5)@—-1) (#—t@+ 2) 
Se eo fa : 


ax—a@ axt2@ #+ar—2’ 
x x—l x—3 


es (2+ cpa ey ae 3)(2b 2) (2a ave ey 


70. 


AAs 





72. 








73. 


Solve: 


80. 


—4[1-40 —2)] 


78. 


9 


at 


tarts 
Gas 2 


war Oe One ate) 
2 (a? — 6’) sree a+b 


GENERAL.REVIEW EXERCISE. 
ee z 
y & 


Ty 


12 
2—3 


tp 
ae as) 


aa 
y 








tee Pt 
74. 


1 75. 











76. 
1—14(1 — 2) 





6a 18. IG: 
ha 25 y 








15 


ere ae 
2(e + 2) 


Qr+a 
3(a% — a) 


x 


ys ay 





359 


360 SCHOOL ALGEBRA. 


20 )°3 4z ihe aaa a 
py) ces ak Amal alk 255 fe) en adie 
a i b c se Ee ie 
4x, 2y> 32 Bi to as 
a0, 4y he 4 here 
a io b C ely Z 
Ba ty 22 5, 2a 6 ¢ 
Carn oN ant, ae 


Find the arithmetical value of 


== 8 


84. 367; 273. 164. 325. 42. 83. 273. 643, 
g5. 32; 64%; Blt; (88)5; (Byy)*; (Ly)! 
86. (0.25)®; (0.027)?; 49°; 32%; 81%, 
87. 3692; 2773. (a) 4; (0.16)-*; (0.0016) * 
88. Interpret a’; a®; a?; (a *)3, 
Simplify : 

1 1 1 1 ‘ad: edhe | 3 po. fe 

BO nL CRC oe ts CS ee Ek ae 


Ne eh Res, ot oa eked Pah eS | yey 
90. .a*b76%Xx a tO Fc $. Obie 4 ap Omar: 


91. (2ab+ 2be+ 2ac— a — 


b? — ¢*) + (a? a bt -[- c?), 


92. V12; V8; V50; V16; 4250; Vi; Wi; VE 
93. 5VW— 820; Vaibi: Va; Varw +a: 


94. 2V18 —38V8 + 2V50; V81+ V24— 


95. 


Rationalize the divisor, and find the value of 








96. : — 98. LX RORB ING 
eS V2 V7 

Oye ek, 99. V8+V2 101. 
38-+V6 V8B=Va 


81\/542°, 
192, 
8/5 + 80 —4-V20; 8-5, + 10-32 — 2V5. 


7V5 


eee ei 


NO 
V3 — V5 


GENERAL REVIEW EXERCISE. 361 





103. 


104, [E44 














105. 


106. ——— = — 





107. 





108. 








109. 


111. 


113. 


1D Qo 

















1 1 tee 
ata b+z2 ab 
7 —_ = ex — bz”, 
Sami hie? aden 
tt+b wxte 
sxy— by = a 
Bay + 382? =22 


e+ 10 ry = spt 
ory —d3y?= 2 


ee var 


kG sf 


ean ae 
Ly — x == 8 


Form the equations of which the roots are: 


116. a—b, a+. 
117. a—26, a+ 30. 
118. at+26, 2a+6. 


Solve: 


122. 2 —5a?+4=0, 
123. 2°— 92° +8=0. 
124. 92*—1827+4=—0. 
125. 42*—1727?+4=0. 
126, -22° — Sai 2 = 0. 


119. 


Lee Ai Sarina \/ 8 


eo Le Noy, lg NS. 
PO UGE ya ia rey Vary 


. 27° — 199% + 24 — 0, 


ieee Pe (); 
ect je): 
Ma Bes = 0). 


Get aT yea! 1 =O) 


SCHOOL ALGEBRA. 


, Qat?—Batt1=0. = 187. a *+5e°4—14=0. | 
. BVe—bV¥2+2=0. 188. 425-3474 27=0, 
. 6V2—8V2—45=0. 139. 2™+32"—4=0. 

01 8/84/2294 S00 (140 Bat = Sorte 
. 8Ve+4VHF—20=0, 141. V2x—V22—2=0. 


142. 8V927+4V32—39=0. 

143. V3ax+av3ax—2a°=0. 

144, 3V2bz —5bW2bc —20°=0. 

145. V2 +44 V382+1=V92e+4. 

146. Vbe+1+2V42—38=10V2—2. 
147. IVe+2—8V82—54+ Vbe41=—0. 
148. V1l—a2+ V8—22—V21+4+22=0. 


ey 


149. 


150. 


151. 


152. 


153. 


. Expand to five terms 5 


( mi 2atyV8ay' (VaF tat (Lea) 


Expand to four terms 
(1—82)*; (1—4a%y?; A—gah?, (a—2274)>. 
Find the eighty-seventh term of (22 —y)™. 


Resolve into partial fractions 


3—22 3—22 ea 
1—824+22?’ (l—2)(1—82)’ 1-# 
3 — 22 


— 32+ 222 











512.9W48S1896 Co01 
A SCHOOL ALGEBRA BOSTON 


7080992 


3H 





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